SHOP  MATHEMATICS 


BY 

EDWARD  E.  HOLTON 

HEAD  OF  DEPARTMENT  OF 
MACHINE  SHOP  PRACTICE 
THE  TECHNICAL  HIGH  SCHOOL 
SPRINGFIELD,  MASSACHUSETTS 


THE  HAYTOL  SERIES  OF 

TEXT  BOOKS  FOR  INDUSTRIAL  EDUCATION 
EDITED  BY  FRANK  E.  MATHEWSON 


THE  TAYLOR-HOLDEN   COMPANY 
SPRINGFIELD,  MASS. 1910 


Copyright,   1910 

BY 

THE  TAYLOR-HOLDEN  COMPANY 

Springfield,  Massachusetts 

Second  Edition 


THE  F.  A.  BASSETTE   COMPANY 

PRINTERS 
SPRINGFIELD,   MASSACHUSETTS 


PREFACE 


This  book  is  the  result  of  twelve  years'  experience  as 
draftsman  and  shop  foreman  combined  with  an  equal  length 
of  service  as  instructor  in  day  and  evening  classes  in  technical 
and  trades  schools. 

The  main  feature  of  the  book  is  the  collection  of  practical 
shop  problems,  the  most  of  them  being  either  actual  prob- 
lems which  have  arisen  in  the  author's  experience  or  those 
suggested  by  that  experience,  and  were  first  collected  for 
use  in  his  own  classes. 

The  formulas  given  are  those  usually  found  in  mechanics' 
hand  books,  and  the  author  acknowledges  his  indebtedness 
to  Wm.  Kent  and  P.  Lobbin  for  permission  to  use  their 
formulas;  also  to  Brown  &  Sharpe  Mfg.  Co.  for  courtesies 
extended,  and  especially  to  Mr.  C.  S.  Bragdon  of  the  Tech- 
nical High  School,  Springfield,  Massachusetts,  for  assistance 
in  collecting  and  arranging  material. 

EDWARD  E.  HOLTON. 
May,  1910. 


2066132 


CONTENTS 


Introduction         .......          vii 

Signs,  Symbols,  Abbreviations,  Etc.  .                    .            ix 

Mechanics             .          .          .          .  .          .          .             1 

Moments  of  forces,  work,  horse  power,  energy,  etc.        1 

Mechanical  Powers        ......  6 

General  law  for  machines         ....  6 

The  six  elements  of  machines  .          .                       6 

The  lever      ....  .                       6 

Compound  lever  9 

Wheel  and  axle                .  10 

Combinations  of  wheels  and  axles  .                     12 

The  pulley              .          .  13 

Fixed  and  movable  pulleys      .  14 

Rule  for  pulleys     .          .  15 

Lathe  test  indicator  18 

The  "  gear  "  of  a  bicycle  21 

Inclined  plane        .                    .  23 

Screw            .  24 

Differential  screw            .  25 

The  wedge  26 

Screw  Threads  29 

Spur  Gears  32 

Bevel  Gears          .  37 

Worm  Gears        ....  .                    41 

Spiral  Gears                             .  .                     42 

Pulleys                                                .  .           47 


CONTENTS 

Friction       .... 
Coefficient  of  friction 
Laws  of  friction     . 
Angle  of  friction 
Leather  Belting 
Manila  Ropes 

Wire  Cables          .... 
Factor  of  Safety 
Chain  Transmission 
Shafting      .  .          . 

Journal  Bearings 
Ball  Bearings 

Machine  Keys      .... 
Linear  Measuring  Instruments 
Machines  .... 

Lathes  ..... 

The  back  gears 
Screw  thread  cutting 
The  reversing  feed  gears 
Compound  gearing 
Turning  tapers 

Cutting  speeds  for  various  materials 
The  metal  planer 

Speeds  and  feeds 
The  Universal  Milling  Machine 
The  index  head 
Compound  indexing 
Differential  indexing 
The  spiral  head 
Speeds  and  feeds  of  cutters 
Drill  press 

Speeds  and  feeds     .          .          . 


vi  CONTENTS 

PAGE 

Hammer  Blow     .                              .....  130 

Horse  Power  of  Machines      .          .          .  132 

The  Prony  brake  135 

Fly  Wheels  138 

Power  of  the  Steam  Engine             .                              .  140 

Power  of  the  locomotive          .          .          .  143 

Power  of  gasolene  engines       .          .                    .  145 

Engine  Cylinders           ......  147 

Indicator  diagram           .....  149 

Steam  Boilers  •  ..          .          .          .          .          .          .  149 

The  H.  P.  of  boilers  150 

To  find  heating  surface  of  shell  and  tubes  151 

To  find  water  capacity             .          .          .          .  152 

To  find  steam  capacity             ...  153 

To  find  pressure  carried  on  stay  bolts       .  153 

To  find  the  bursting  strength            .  154 

The  safety  valve              .          .          .          .          .  157 

Hydraulics            .          .          .          .          .          .          .  160 

Pressure  and  head           .          .  160 

Archimedes  discovery               .          .          .          .  161 

Specific  gravity      ...                    .          .  161 

Hydraulic  machines        .          .          .          .          .  163 

Tables  of  Standard  Units  of  Weights  and  Measures    .  169 

The  Metric  System        .  173 

Table  of  Decimal  Equivalents         .  177 

Use  of  Formulas           ...  178 

Simple  Trigonometric  Functions     .  186 

Table  of  Natural  Sines,  Etc.                              .  189 

Formulas              .......  199 

Answers      ........  203 

Index  221 


INTRODUCTION 

CHARLES  F.  WARNER,  Sc.  D. 

Principal  of  the  Technical  High  School  and  of  the  Evening  School 
of  Trades,  Springfield,  Mass. 


The  loudest  note  sounded  in  all  recent  educational  dis- 
cussions is  the  call  for  more  practical  methods  in  presenting 
the  fundamental  subjects  of  study.  It  is  claimed  that 
language,  science,  and  mathematics  have  been  taught  with 
too  little  reference  to  their  utility  in  the  vocations  and  in  the 
ordinary  affairs  of  life.  The  methods  commonly  employed  in 
presenting  mathematical  subjects  have  been  especially  open 
to  criticism.  This  is  true  even  in  some  schools  designed  to 
give  vocational  training.  Mathematical  books  of  the  kind 
traditional  in  the  older  schools  have  been  continued  in  use 
in  the  newer  and  more  practical  schools.  These  books  were 
written  almost  entirely  from  the  point  of  view  of  the  teacher 
of  pure  mathematics  with  little  reference  to  concrete 
problems  of  life  and'  having  no  reference  whatever  to  the 
actual  problems  of  the  drafting-room  and  the  shop.  As  a 
natural  consequence  the  class  room  work  in  mathematics, 
in  many  of  our  most  practical  schools,  has  failed  to  utilize 
the  material  afforded  by  the  shops  and  science  laboratories 
to  fix  the  knowledge  of  mathematical  principles  by  concrete 
illustration  and  by  practice. 

There  is  much  truth  in  the  criticism.  But  what  are  we 
going  to  do  about  it?  It  will  not  do  to  cast  aside  the  old- 
time  algebras  and  geometries  unless  something  really  better 
can  be  found  to  take  their  place.  The  effort  to  make  the 


viii  SHOP  MATHEMATICS 

applications  of  mathematics  more  easily  understood  might 
lead  to  the  substitution  of  a  practical  course  in  which  the 
mathematical  element  is  too  much  diluted.  This  would  be 
folly.  The  real  object  should  be  to  strengthen, — not  to 
weaken — the  teaching  of  mathematics  in  practical  schools. 
What  is  needed  is  to  purge  the  old  books  of  useless  material 
and  put  in  place  of  it  practical  mathematical  work  distinctly 
planned  to  make  up  for  the  short-comings  of  the  old  methods 
when  measured  by  the  practical  demands  of  modern  times. 
The  author  of  "Shop  Mathematics"  has  had  many 
years'  experience  in  designing  and  making  machine  tools  and 
in  a  wide  range  of  practical  shop  work.  In  addition  to  this 
he  has  had  a  long  experience  as  a  teacher  of  drawing  and  of 
machine  shop  practice  and  tool-making  in  schools  for  boys 
and  for  adults.  This  has  given  him  an  unusual  opportunity 
not  only  to  find  out  what  is  needed,  but  to  discover  the 
facilities  for  supplying  that  need.  The  book  contains  a 
selection  of  problems  that  actually  arise  in  shop  practice. 
This  is  what  is  needed  by  the  young  mathematical  student 
and  for  two  reasons, — first,  that  he  may  know  what  the  shop 
problems  are,  and,  second,  that  he  may  learn  how  to  apply 
mathematical  principles,  rules,  and  formulas  in  the  solution 
of  such  problems.  No  attempt  is  made  in  this  book  to  teach 
mathematical  theory  or  principles.  That  would  be  a  need- 
less repetition  of  countless  books  already  in  existence.  "Shop 
Mathematics"  may  be  used  to  supplement  the  course  in 
elementary  algebra,  geometry,  and  trigonometry,  and  if 
used  in  this  way  it  will  be  found  of  great  value  in  technical 
schools.  But  abundant  rules  and  formulas  are  given  under 
each  subject  so  that  the  book  will  also  find  a  place  in  brief 
practical  courses  which  do  not  admit  of  the  use  of  the 
ordinary  mathematical  text-book. 


SIGNS,  SYMBOLS,  AND  ABBREVIATIONS 
USED  IN  THIS  BOOK 


+  ,  plus,  is  the  sign  of  addition. 

— ,  minus,  is  the  sign  of  subtraction. 

Plus  and  minus  are  also  used  to  indicate  positive  and 
negative  quantities. 

X ,  times,  or  multiplied  by,  is  the  sign  of  multiplication. 
A  dot  .  is  sometimes  used  for  X  where  the  quantities  are 
expressed  by  letters,  but  is  usually  omitted  in  algebraic 
formulas;  thus  a  X  b,  or  a  .  6  is  ordinarily  written  ab. 

H-  divided  by,  is  the  sign  of  division. 

:  without  the  dash  between  also  indicates  division,  being 
used  as  the  ratio  sign;  as,  a  :  b,  means  the  ratio  of  a  to  6, 
or  a  divided  by  b. 

=  is  the  sign  of  equality  and  indicates  that  the  two 
quantities  between  which  it  is  placed  are  of  equal  value. 

.'.,  therefore. 

>,  greater  than;   as,  6>4,  read  6  is  greater  than  4- 

<,  less  than;    as,  4<6,  read  4  is  less  than  6. 

Z ,  angle. 

J_,  perpendicular  to;  as  A±_B,  read  A  is  perpendicular 
to  B. 

||,  parallel  to;  as,  C  \\  D,  read  C  is  parallel  to  D. 

.,  the  decimal  point;  as,  0.2,  read  two  tenths;  or  0.004, 
four  thousandths. 


x  SHOP  MATHEMATICS 

°,  the  symbol  for  the  degree  in  the  measurement  of  angles. 

'  and  ",  the  accents,  denote  minutes  and  seconds  in  the 
measurement  of  angles;  as,  5°  10'  15" ,  read  5  degrees  10 
minutes  15  seconds. 

The  above  symbols,  '  ",  are  also  used  for  feet  and  inches 
in  indicating  dimensions;  as,  3'  6" ,  read  3  feet  6  inches. 

The  subscript  is  a  small  figure  written  at  the  lower  right 
of  a  letter;  as,  Alt  A2,  A3,  read  A  sub  one,  A  sub  two,  A 
sub  three.  It  is  used  to  denote  corresponding  parts  of 
related  objects,  or  sometimes  to  avoid  using  too  many 
different  letters. 

\/,  the  radical  sign,  denotes  the  square  root. 

Other  roots  are  indicated  by  a  small  figure,  called  the 
index,  written  at  the  upper  left  of  the  radical  sign;  as, 

3         5         n 

\/,  \/,  v7,  read  cube  root,  fifth  root,  nth  root. 

The  exponent  is  a  small  figure  written  at  the  upper  right 
of  a  quantity  to  indicate  a  power,  or  the  number  of  times  the 
quantity  is  used  as  a  factor;  as,  52  means  5X5,  3*  = 
3X3X-3X3. 

(  ),  parentheses,  {  }  braces,  [  ]  brackets, vinculum, 

signify  that  the  inclosed  quantities  are  to  be  considered 
as  one  quantity. 

sin,  sine. 

cos,  cosine. 

tan,  tangent. 

cot,  cotangent. 

sec,  secant. 

cosec,  or  esc.,  cosecant. 

TT,  Pi,  the  ratio  of  the  circumference  of  a  circle. to  its 
diameter,  =3.1416. 

D,  or  dia.,  diameter  of  a  circle. 


SHOP  MATHEMATICS  xi 


R,  or  r,  radius  of  a  circle. 

P  or  F,  power  or  force. 

W,  weight  or  resistance. 

H.  P.,  horse  power. 

R.  P.  M.,  revolutions  per  minute. 

F.  P.  M.,  feet  per  minute. 

/,  coefficient  of  friction. 

ft.-lb.,  foot  pound. 

ft.,  foot  or  feet. 

Z&.,  pound. 

in.,  inch  or  inches. 

pi.,  pitch. 


MECHANICS 

Mechanics  treats  of  forces  and  of  the  effects  of  forces. 
Force  is  the  action  between  two  bodies  tending  to  produce  a 
change  of  position  or  shape;  as  when  a  horse  pulls  a  load,  a 
motor  drives  an  electric  car,  elasticity  causes  the  action  of  a 
steel  spring,  etc. 

The  moment  of  a  force.    If  a  bar  of  uniform  size  is  pivoted 
at  its  center  and  a  weight  placed  on  one  end,  the  bar  will 
rotate  about  the  pivot.     The  numerical  value  of  the  im- 
portance of  the  force  in  producing  motion  about  a  pivot 
is  called  its  moment  and  is  equal  to  the  product  of  the  force 
by  the  distance  from  its  line  of  action  to  the  pivot.     By 
reference  to  the  accompanying  sketch,  the  moment  of  the 
force  FI,  applied  at  point  A,  is  seen  to  be  6X  10  =  60  ft.Jbs. 
of  F2,  4%  X10  =  45  ft.-lbs. 
of  F3,  3    X 10  =  30  ft.-lbs. 

Diagram  for  Momenta  of  Force 

,/>. 


When  a  force  acting  upon  a  body  changes  its  position, 
work  is  done  upon   the  body.     The  amount  of  work  done 


2  SHOP  MATHEMATICS 

depends  upon  the  force  applied,  also  upon  the  distance 
through  which  it  acts;  that  is,  work  is  measured  by  the 
resistance  overcome,  multiplied  by  the  distance  through 
which  it  is  overcome;  as  a  10  pound  weight  that  is  lifted 
to  a  height  of  5  feet  requires  an  amount  of  work  equal 
to  the  product  of  10  times  5  =  50  ft.-lbs. 

The  unit  of  measure  in  the  above  example  is  expressed 
by  the  term  foot-pound,  that  is,  a  force  of  one  pound  acting 
through  a  distance  of  one  foot,  or  its  equivalent;  as  4 
pounds  acting  through  a  distance  of  \  foot,  or  TV  pound 
acting  through  10  feet,  etc. 

The  unit  can  be  expressed  not  only  in  ft-lbs.  but  in  inch- 
pounds,  foot-tons,  centimeter-grams,*  etc. 

The  amount  of  work  done  in  lifting  a  body  a  given  distance 
is  the  same  whether  clone  in  5  seconds  or  5  minutes,  but  it 
is  often  necessary  also  to  denote  the  rate  of  doing  work. 

This  is  expressed  in  terms  of  horse  power.  An  engine  of 
one  horse  power  (1  H.  P.)  means  an  engine  capable  of 
doing  33,000  ft.-lbs.  of  work  in  one  minute. 

Therefore  to  find  the  horse  power  of  any  machine  the 
formula  used  is, 

_     Ft.-lbs.  of  work  done 
33000  X  time  in  minutes 

In  electric  power  machines  such  as  dynamos  and  motors 
one  H.  P.  is  equal  to  746  watts;  then  the  formula  for  power 
of  electric  machines  is: 

volts  X  amperes 

~r46~~ 

Energy  is  the  capacity  for  doing  work;  as  a  coiled  spring 
has  the  capacity  to  set  in  motion  the  mechanism  of  a  clock 

*Note.  For  dimensions  in  the  metric  system  see  tables,  pages 
173  to  176. 


SHOP  MATHEMATICS  3 

or  watch.  The  coiled  spring  of  a  watch  possesses  energy 
because  at  some  previous  time  work  has  been  performed 
upon  it  in  the  winding. 

The  amount  of  energy  in  a  body  is  measured  in  ft.-lbs., 
the  unit  used  for  measuring  work.  Mechanical  energy  is  of 
two  kinds,  potential,  and  kinetic.  Potential  energy  is  due  to 
the  position  of  a  body;  as,  for  example,  if  a  pile  driver  head 
weighing  50  pounds  is  suspended  20  feet  above  the  ground  , 
it  has  a  potential  energy  of  20X50  =  1000  ft.-lhs.  If  now 
the  weight  is  released  and  falls,  the  energy  is  of  a  different 
kind  because  of  the  motion  of  the  falling  weight.  This  is 
called  kinetic  energy.  In  either  case  the  weight  times  the 
height  equals  the  measure  of  the  energy  of  the  body  or 
K  =  Wh. 

When  the  velocity  of  the  falling  body  is  given  instead  of 
the  height  from  which  it  falls, 

Then  by  the  laws  of  falling  bodies 

' 


Where          v  =  velocity  in  feet  per  second 
g  =  32.16  feet 

Wv2 

Then          K  =  ^- 
2g 

K  =  the  kinetic  energy 

W  =  weight  of  the  body  in  pounds. 

PROBLEMS 

1.  A  drop  hammer  weighing  400  Ibs.  falls  from  a  height 
of  36  in.    What  kind  and  how  much  energy  will  be  exerted? 

2.  A  weight  of  500  Ibs.  is  used  for  breaking  up  old  car 
wheels  and  is  suspended  15  ft.  above  the  anvil  block.    Cal- 
culate the  kind  and  amount  of  energy. 


4  SHOP  MATHEMATICS 

3.  An  elevator  car  weighing  2  tons  requires  how  much 
energy  to  lift  it  a  distance  of  20  feet? 

4.  A  tank  contains  5,000  gals,  of  water,  50  ft.  being  the 
average  height  above  the  ground.     What  is  the  energy  of 
the  water  at  the  ground? 

5.  A  pile  driver  head  weighing  175  Ibs.  falls  from  a 
height  of  18  ft.    What  is  the  energy  at  the  end  of  the  fall? 

6.  A  man  weighing  180  Ibs.  jumps  up  to  a  platform 
42  inches  above  the  ground.     How  much  energy  was  ex- 
pended? 

7.  If  the  man  in  problem  6  should  climb  a  distance 
of  100  ft.  above  a  certain  point,  how  much  energy  would 
be  exerted? 

8.  If  a  hammer  that  weighs  1  Ib.  has  a  velocity  of  22 
ft.  per  sec.,  what  is  its  energy? 

9.  If  an  iron  ball  weighs  100  Ibs.,  what  is  its  energy 
when  suspended  at  a  height  of  28  ft.  from  the  ground? 

10.  A  rock  drill  is  operated  by  a  10  Ib.  sledge  hammer 
with  a  striking  velocity  of  30  ft.  per  sec.     Find  the  energy 
used  in  the  drilling. 

11.  W'hat  is  the  H.  P.  of  a  pump  that  raises  100,000  gals, 
of  water  per  hour  to  a  reservoir  25  ft.  above  the  level  of  the 
lake  from  which  it  is  pumped? 

Note.     One  cu.  ft.  of  water  weighs  approximately  62£  Ibs.  and 
contains  about  1\  gals. 

12.  A  motor  is  able  to  lift  an  elevator,  which  with  load 
weighs  5  tons,  to  the  top  of  a  tower  500  ft.  high  in  two  min- 
utes.    What  H.  P.  is  required  to  do  the  work? 

13.  What  must  be  the  H.  P.  of  the  engine  required  to 
raise  a  block  of  granite  weighing  10  tons  to  the  top  of  a 
wall  35  ft.  from  starting  point  when  it  takes  20  minutes 
to  do  the  work? 


SHOP  MATHEMATICS  5 

14.  If  it  takes  1^  hours  to  raise  a  weight  of  20  tons  100  ft., 
what  H.  P.  engine  will  be  required? 

15.  If  a  workman  carries  5  tons  of  pig  iron  up  a  flight 
of  steps  14  ft.  high  in  10  hours,  how  much  work  does  he 
accomplish  expressed  in  ft.-lbs.? 

16.  A  rope  turns  a  pulley  48  in.  dia.  with  a  pull  of  10 
Ibs.  at  the  rim  at  the  rate  of  2500  ft.  per  minute.    How  many 
ft.-lbs.  of  work  are  done  in  5  hours  consecutive  movement? 

17.  A  12  Ib.  ball  falls  2500  ft.    What  is  its  velocity  when 
it  strikes  and  what  is  its  kinetic  energy? 

18.  A  500  volt  electric  motor  runs  50  machines  using  22 
amperes  of  current.     What  is  its  H.  P.? 

19.  What  is  the  H.  P.  of  a  dynamo  which  will  run  300 
110-volt  incandescent  lamps  if  each  lamp  consumes  ^  ampere 
of  current? 

20.  An  electric  motor  has  a  voltage  of  250  and  supplies 
30  amperes  to  run  a  certain  set  of  machines.     Find  the 
H .  P.  of  motor. 

21.  What  H.   P.  will  be  required  to   run  a  250  lamp 
circuit,  the  lamps  being  the  same  as  in  problem  19? 

22.  A  direct  connected  dynamo  delivers  power  at  110 
volts  to  a  1400  electric  light  circuit,  also  to  9  motors  with 
normal  amperage  as  follows:    1  at  144,  1  at  110,  1  at  80, 
3  at  76,  1  at  57  and  2  at  20  amperes.    What  H.  P.  will  be 
required  to  run  the  above  equipment? 


MECHANICAL  POWERS 

An  appliance  by  which  force  can  be  used  to  do  useful 
work  is  called  a  machine. 

The  mechanical  powers  or  elements  of  machines  are  six 
in  number,  as  follows: 

1.  Lever. 

2.  Wheel  and  Axle. 

3.  Pulley. 

4.  Inclined  plane. 

5.  Screw. 

6.  Wedge. 

The  mechanical  advantage  of  all  kinds  of  machines, 
whether  simple  or  compound,  may  be  computed  in  accordance 
with  the  following 

GENERAL  LAW 

The  force  multiplied  by  the  distance  through  which  it 
moves  is  equal  to  the  resistance  or  weight  multiplied  by  the 
distance  through  which  it  moves,  or  P  X  dP  =  W  X  dW. 

Each  class  of  machines  permits  of  a  special  statement  of 
this  law  by  substituting  for  the  general  terms  the  special 
terms  used  with  that  class  of  machines. 

1.    THE  LEVER 

The   lever  is  an  inflexible  bar  or  rod  supported  at  some 
point,  the  bar  being  free  to  move  about  that  point  as  a  pivot. 
This  pivotal  point  is  the  fulcrum,  usually  represented  by  F. 
The  force  applied  to  the  lever  is  represented  by  P. 


SHOP  MATHEMATICS 


The  weight  lifted,  or  the  resistance  to  the  force  is  repre- 
sented by  W. 

The  lever  is  classed  according  to  the  position  of  the  three 
points  P,  F,  W. 

A  lever  of  the  first 
class,  Fig.  1,  has  F 
between  P  and  W. 


A  lever  of  the  second 
class,  Fig.  2,  has  W 
between  F  and  P. 


A  lever  of  the  third 
class,  Fig.  3,  has  P 
between  F  and  W. 


The  machinists'  and  the  tinsmiths'  pliers  are  -examples 
of  levers  of  the  first  class. 

The  nut  cracker  and  lemon  squeezer,  are  levers  of  the 
second  class. 

The  sheep  shears  and  firm  joint  calipers  are  good  examples 
of  levers  of  the  third  class. 

The  three  classes  of  levers  are  operated  and  controlled  by 
the  following: 

LAW  FOR  LEVERS.  The  force  multiplied  by  its  distance 
from  the  fulcrum  is  equal  to  the  weight  multiplied  by  its 
distance  from  the  fulcrum. 


SHOP  MATHEMATICS 


Let     P  =  force 

W  =  weight  or  resistance. 

Pa  =  distance  from  fulcrum  to  point  where  force  is 

applied. 
Wa=  distance  from  fulcrum  to  point  where  weight  is 

applied. 

Then  the  law  of  Levers  becomes  PxPa  =  WxWa. 
From  this  equation  the  following  are  readily  derived. 


WxWa 


W  = 


PxPa 


WxWa 


PXPa 


Pa  Wa  P  W 

Example.     What  force  18  in.  from  fulcrum  will  balance  a 
weight  of  870  Ibs.  3  in.  from  the  fulcrum? 

Solution.     By  formula: 

WxWa  870X3 


P  = 


Pa 


6 


=  145  Ibs.,  Ans. 


The  law  for  bent  levers  is  the  same  as  for  straight  levers 
but  the  lengths  of  arms  are  computed  on  lines  from  the 
fulcrum  at  right 
angles  to  -the  direc- 
tions  in  which  P 
and  W  act. 

The  lengths  of  the 
arms  of  a  bent  lever 
are  continually 
changing  as  the 
force  and  weight 
move  into  new 
positions.  The  case  of  pulling  out  a  nail  with  the  common 
claw  hammer  is  a  good  illustration  of  the  bent  lever,  Fig.  4. 


SHOP  MATHEMATICS 


^^^$5^^ 


The  moving  strut  in  Fig.  5  is  a  bent  lever  with  one  arm 
lacking.  The  force  is 
applied  at  the  same  end 
at  which  the  resistance 
is  to  be  overcome.  The 
resistance  in  this  case  is 
not  the  weight,  W,  but  its  resistance  to  being  moved.  The 
ratio  between  force  and  resistance  changes  as  the  angle 
A  changes. 

Then 

force:  resi stance  =  sin  A:  cos  A 
or 

P  X  cos  A  =  W  X  sin  A 

WXsinA  PXcos  A 

and       P  = -. —     W  = : — ; — 

cos  A     ;  sin  A 

The  toggle    joint   is  a   double 
strut,     Figs.  6  and  7. 
The  statement  is 

P  :  W  =  2  sin  A  :  cos  A 
P_WX2  sin  A 

cos  A 

w_PXcos  A 
2  sin  A 

A  compound  lever  is  a  combi- 
nation of  two  or  more  levers 
which  may  be  of  the  first,  second 
or  third  classes.  The  calcula- 
tions for  compound  levers  can 
be  made  by  taking  each  lever 


' 


separately  by  the  formula  for  single  levers.  These  separate 
operations,  however,  are  usually  condensed  into  one,  in 
accordance  with  the  following: 


10 


SHOP  MATHEMATICS 


LAW  OF  COMPOUND  LEVERS.  The  continued  product 
of  the  force  and  all  the  force  arms  is  equal  to  the  continued 
product  of  the  weight  and  all  the  weight  arms. 

This  law  gives  rise  to  the  following  formulas: 
P  = 


W  = 


PaXPa.XPa.,. 
PxPaXPa.XPa^ 


H.    WHEEL  AND  AXLE 


The  wheel  and  axle  may  be  considered  a  continuous  lever. 
By  its  use  a  continuous  motion  is  obtained  for  raising  a 
weight.    The  two  arms  of  the  lever  are  the  diameters  of  the 
wheel  and  axle,  or  the  radii  of  wheel  and  axle. 
If  D  is  diameter  of  wheel: 
d=  diameter  of  axle 
R  =  radius  of  wheel 
r  =  radius  of  axle 
C  =  circumf erence  of  wheel 
c=  circumference  of  axle 
Then  these  formulas  apply :  P     W  =  d  :  D 

P     W  =  r  :R 
P     W  =  c:C 


Fry.  8, 

Fig.  8  shows  sketch  of  simple  windlass. 

Fig.  9  shows  sketch  of  capstan,  which  is  a  wheel  and  axle 


SHOP  MATHEMATICS 


11 


where  the  axle  has  a  vertical  position.     It  is  used  on  ships 
for  raising  the  anchor. 

The  mechanical  advantage  of  the  wheel  and  axle  may 
be  increased  by  making  the  diameter  of  wheel  larger  or  the 
diameter  of  axle  smaller. 

The  windlass  is  a  modification  of  the  wheel  and  axle  where 
a  crank  is  substituted  for  the  wheel.  It  is  used  especially 
where  the  power  is  applied  by  hand. 

The  differential  windlass,  Fig.  10,  is  a  device  for  increasing 
the  mechanical  advantage  of 
the  axle  by  unwinding  rope 
from  a  small  axle  and  wind- 
ing it  on  to  an  axle  of  larger 
diameter;  for  with  one  turn 
of  crank  C,  the  section  of  rope 
supporting  the  weight  will  be 
shortened  a  distance  equal  to 
circumference  of  the  large  axle 
minus  the  circumference  of  the 
small  axle  and  W  will  be  raised  half  this  distance. 
By  the  law  of  the  wheel  and  axle, 

P  :  W  =  r  :R 

then  in  the  differential  windlass 
(r.-r) 

2  * 

rl  =  radius  of  large  axle 
r  =  radius  of  small  axle 

When  motion  is  transmitted  from  one  body  to  another 
by  direct  or  by  indirect  contact  the  body  that  produces 
the  motion  is  called  the  driver,  the  body  that  receives  the 
motion  is  called  the  follower. 

In  a  combination  of  wheels  and  axles  each  pair  may  be 


12  SHOP  MATHEMATICS 

determined   separately,   but  the  shorter  way  is  to   use  a 
formula  similar  to  that  used  for  compound  levers,  as  follows : 


Where  R,  R1}  R2,  are  the  radii  of  the  wheels  and  r,  i\,  r2, 
are  the  radii  of  the  axles. 

These  problems  may  also  be  calculated  by  the  following: 
RULE  FOR  DRIVERS  AND  FOLLOWERS.  The  speed 
of  the  first  driver  multiplied  by  the  product  of  the  sizes  of 
all  the  drivers  is  equal  to  the  speed  of  the  last  follower 
multiplied  by  the  product  of  the  sizes  of  all  the  followers. 
For  the  sizes  of  followers  and  drivers  may  be  taken  the 
circumferences,  diameters,  radii  or  number  of  teeth  (if  a  gear 
wheel) ;  only  whatever  dimension  is  taken  for  any  driver  the 
corresponding  dimension  must  be  taken  for  its  follower. 
Thus  for  a  train  of  wheels  having  four  drivers  and  followers, 
the  diameters  can  be  taken  for  one  pair,  the  radii  for  another, 
the  circumferences  for  a  third  and  number  of  teeth  for  the 
fourth,  to  solve  the  problem. 

Let  N  =  th&  R.  P  .M.  of  first  driver 
n  =  the  R.  P.  M .  of  last  follower 
D,  Dlf  D2  =size  of  drivers 
F,  Flt  F2  =  size  of  followers 
Then  by  formula, 

N  X  D  X  Dl  X  D2 

F '  X  Fl  X  F2 

The  size  of  any  driver  or  follower  can  be  found  by  the 
formula : 

D_X_N 
*~~        n 

FXn, 
D~      N 

where,  N—R,  P.  M.  of  the  driver 
n  =  R.  P.  M.  of  the  follower 


SHOP  MATHEMATICS 


13 


Figures  12,  13,  14  and  15  are  good  illustrations  of  the  above 
rule  in  shop  practice.  Fig.  12,  shows  a  combination  of  the 
stud  and  change  gears  used  on  the  engine  lathe  for  thread 
cutting,  and  on  the  milling  machine  for  cutting  spiral  flutes 
in  cutting  tools.  Fig.  13,  shows  the  belting  from  main  line 
drive  pulley  through  two  countershafts  to  spindle  pulley  of 
machine.  Fig.  14  shows  sketch  of  chain  and  sprocket  drive 
for  bicycle  or  automobile,  which  form  of  drive  is  also  being 
used  extensively  for  positive  drives  in  some  kinds  of  machine 
construction.  Fig.  15  is  the  bevel  gear  drive. 

III.     THE  PULLEY 

The  pulley  is  a  wheel  over  which  a  cord,  band  or  chain  is 
passed  to  transmit  the  force  applied  to  the  cord  in  another 
direction.  The  wheel  is  introduced  to  diminish  the  friction, 


14 


SHOP  MATHEMATICS 


the  band  being  the  part  that  gives  practical  effect  to  the 
machine. 

There  is  no  mechanical  advantage  gained  with  the  fixed 
pulley  as  shown  in  Fig.  16,  but  as  stated 
above  it  is  of  great  use  in  changing  the 
direction  of  the  force. 

The  fixed  pulley  when  used  in  combi- 
nation with  the  movable  pulley  as  shown 
at  B  in  Figs.  17  and  18  has  mechanical 
advantage,  since  the  weight  is  carried 
by  strands  of  the  cord  on  either  side 
of  the  movable  pulley. 
The  usual  arrangement  of  pulley  blocks  is  shown  in  Figs. 

17   and    18   where   one   or 

more  wheels  or  sheaves 
(grooved  pulleys)  are 
placed  in  suitable  pivots 
and  bearings,  and  cords  are 
passed  over  the  pulleys 
connecting  the  force  and 
weight. 

In  Fig.  17,  if  there  is  a 
single  pulley  at  A  and  B 
and  a  rope  is  passed  over 
A  and  around  B  and  fas- 
tened at  C,  then  a  pull  of  1 
pound  at  P  will  transmit  a 
pull  of  1  pound  to  the  rope 
on  the  other  side  of  pulley  A ; 
this  will  transmit  the  same 
amount  of  force  at  B,  which  in  turn  will  transmit  the  same 
force  to  the  other  side  of  pulley  B.  Thus  1  pound  pull  at  P 


SHOP  MATHEMATICS 


15 


will  balance  1  pound  at  1  and  2  or  the  combined  puil  of  1 
and  2  at  W,  so  that  1  pound  at  P  will  support  2  pounds  at  W. 
Again,  if  rope  at  P  moves  down  1  foot  the  rope  must  move 
up  1  foot  on  the  other  side  of  A:  but  the  end  of  rope  is 
fixed  at  C ' ,  so  that  when  rope  moves  up  at  1  and  remains 
stationary  at  2,  one-half  the  motion  will  be  given  to  the 
movable  pulley  between  1  and  2,  which  is  in  accordance 
with  the  general  law  of  machines.'. W  =  2  P. 

If  there  are  two  sheaves  on  each  block  A  and  B,  each 
turning  independently  of  the  other  in  the  bearings,  the  pull 
of  1  pound  at  P  will  be  transmitted  as  in  the  first  case  and  will 
also  transmit  the  pull  to  the  third  and  fourth  strands,  so 
that  a  pull  of  1  pound  at  P  will  balance  a  weight  of  4  pounds 
at  W.  Similarly  it  can  be  shown  that  1  foot 
downward  travel  at  P  will  give  an  upward 
travel  of  \  foot  at  W.  From  these  prin- 
ciples is  obtained  the  following: 

RULE  FOR  PULLEYS.  The  force  multi- 
plied by  the  number  of  strands  from  the 
movable  pulley  will  equal  the  weight  that 
can  be  raised, 

or  PXN=W, 

where  N  is  the  number  of  strands  from  the 
movable  pulley. 

Whenever  possible  the  pulleys  should  be 
arranged  so  as  to  pull  in  the  direction  in 
which  the  weight  is  to  be  moved,  as  shown 
in  Fig.  18,  for  whereas  in  Fig.  17,  N  =  2, 
in  Fig.  18,  N  =  3. 

The  differential  pulley,  Fig.  19,  is  quite  generally  used  in 
shop  practice,  the  band  for  transmitting  the  force  being  an 


16  SHOP  MATHEMATICS 

endless-chain  of  iron  links.  The  principle  of  the  differential 
pulley  is  very  similar  to  that  of  the  differential  windlass. 
The  radius  of  the  movable  pulley  F  must  be  an  exact  mean 
between  the  radii  of  B  and  C,  in  order  to  have  a  uniform 
velocity  ratio,  so  that  chain  links  may  fit  the  pockets  in 
sprocket  wheels. 

Let  R  =  radius  of  pulley  B, 
r  =  radius  of  pulley  C. 

While  the  pull  at  P  moves  the  chain  between  B  and  D  up 
a  distance  R,  the  chain  at  C  will  move  down  a  distance  r, 
the  loop  around  the  pulley  D  will  be  shortened  the  distance 
R-r  and  W  will  be  raised  one-half  this  amount.  From  this 
is  obtained  the  statement  as  follows: 

PXR=WXR~T 


WX 


2 

R-r 

2 


w= 


R 

PXR 
R-r 


MISCELLANEOUS  PROBLEMS 

In  the  following  problems  the  weight  of  the  lever  and 
friction  of  the  bearings  will  not  be  considered  in  the  cal- 
culations. 

1.  What  force  36  in.  from  the  fulcrum  will  balance  a 
weight  of  500  Ibs.,  9  in.  from  the  fulcrum  in  a  lever  of  the 
first  class? 


SHOP  MATHEMATICS  17 

2.  If  a  pull  of  75  Ibs.  is  made  on  the  end  of  a  bar  6  ft. 
long  with  the  fulcrum  6  in.  from  the  other  end,  what  weight 
in  a  lever  of  the  first  class  will  just  balance  the  pull? 

3.  If  the  lever  of  problem  2  had  been  of  the  second  class 
and  the  weight  6  in.  from  the  end,  what  weight  would  have 
been  required? 

4.  When  a  weight  of  324  Ibs.  is  balanced  on  the  end  of 
a  lever  of  first  class  by  a  force  of  62  Ibs.  27  in.  from  fulcrum, 
what  distance  is  the  weight  from  fulcrum? 

5.  When  a  weight  of  685  Ibs.  2£  in.  from  the  fulcrum 
is  balanced  on  a  bar  by  a  pull  of  75  Ibs.,  what  distance  is 
the  force  from  the  fulcrum? 

6.  What  weight  4£  in.  from  the  fulcrum  will  be  balanced 
on  a  bar  by  a  force  of  96  Ibs.  48  in.  from  the  fulcrum? 

7.  What  force  10  ft.  from  the  fulcrum  will  just  raise  a 
weight  of  2465  Ibs.  8  in.  from  the  fulcrum  with  a  lever  of 
the  second  class? 

8.  A  bar  of  iron  was  placed  with  one  end  against  the 
wall  of  a  shop,  the  other  end  being  placed  against  the  leg  of 
a  machine  that  weighed  1500  Ibs.     If  the  bar  lay  at  an 
angle  of  10°  to  the  floor,  what  weight  would  be  required  on 
the  end  next  to  the  machine  to  move  the  machine  along  the 
floor  assuming  that  the  resistance  is  equal  to  the  weight  of 
the  machine? 

9.  A  quartz  crusher  jaw  requires  a  force  of  10,000  Ibs. 
to  crack  a  certain  stone.    If  the  angle  through  which  toggle 
arms  move  is  5£°,  find  the  force  that  must  be  put  on  the 
toggle  arms  to  break  the  stone. 

10.     The  power  on  the  end  of  a  strut  is  124  Ibs.,  the  angle 
is  5°.     What  resistance  could  it  overcome? 


18  SHOP  MATHEMATICS 

11.  A  baling  press  requires  a  pressure  of  6000  Ibs.  to  bale 
a  bundle  of  steel  chips.    The  platen  is  operated  by  a  toggle 
joint  with  arms  at  an  angle  of  10°.    What  force  will  be  re- 
quired to  operate  the  press? 

12.  The  knife  of  a  paper  shear  is  operated  by  a  toggle 
joint,  the  arms  beginning  the  cut  at  an  angle  of  4^°.     If  a 
pull  of  650  Ibs.  is  required  to  start  the  arms  to  obtain  a  cut, 
what  is  the  resistance  on  the  knife  blade? 

Fig.  20  shows  a  practical  example  of  a  lever  of  the  first 
class.  The  long  end  of  the  pointer  at  C  multiplies  the  error 
in  the  position  of  point  A  with  respect  to  the  axis  of  the 
lathe  spindle,  in  proportion  to  the  length  of  the  arms  A  B 
and  B  C.  If  the  distance  from  A  to  B  is  \  inch  and  from 
B  to  C  is  8  inches  the  amount  of  motion  of  C  will  be  in  the 
ratio  of  8  to  or  16  tunes  as  much  as  A. 


13.  The  pointer  of  a  lathe  test  indicator  is  15  in.  long 
over  all;  if  the  short  arm  of  lever  is  .375  in.  and  point  c 
moves  in  a  path  ^  in.  across,  what  amount  of  error  in  posi- 
tion of  the  point  A  is  indicated? 


SHOP  MATHEMATICS  19 

14.  The  bar  of  a  lathe  test  indicator  is  13  in.  from  the 
end  to  a  point  in  the  work  to  be  tested  and  the  short  arm  is 
^  in.    How  large  a  circle  will  be  described  at  long  end  when 
short  end  is  ^^^  in.  out  of  line  with  the  axis  of  lathe  spindle? 

15.  The  axis  of  a  windlass  is  6  in.  dia.  and  the  crank  is 
20  in.  long.     If  a  pull  of  56  Ibs.  is  given  at  the  end  of  the 
crank  what  weight  will  be  raised? 

16.  Six  men  each  exert  a  force  of  125  Ibs.  on  the  ends  of 
3  ft.  capstan  bars,  the  barrel  of  capstan  is  16  in.  and  the 
bars  enter  6  in.  into  the  cap  of  capstan,  which  is  24  in.  dia. 
What  weight  will  be  lifted? 

17.  The  steering  wheel  of  a  ship  is  6  ft.  dia.  and  the  drum 
is  12  in.  dia.     What  resistance  can  be  overcome  on  ropes 
from  drum  by  a  force  of  200  Ibs.  at  the  rim  of  the  steering 
wheel? 

18.  The  crank  of  a  differential  windlass  is  30  in.  long  and 
the  force  applied  at  the  end  of  the  crank  is  75  Ibs.     When 
the  diameters  of  the  axles  are  12  in.  and  14  in.,  what  weight 
will  be  raised? 

19.  A  weight  of  10,000  Ibs.  is  to  be  raised  by  a  differential 
windlass  with  a  force  of  100  Ibs.    What  length  of  crank  will 
be  required  with  axles  12  in.  and  13  in.  diameters? 

20.  What  is  the  force  required  on  a  crank  18  in.  long  on 
a  differential  windlass,  with  axles  18  in.  and  20  in.  diameters, 
to  raise  a  weight  of  8,000  Ibs? 

21.  A  compound  lever  of  first,  second  and  third  classes 
has  levers  each  4  ft.  long,  the  short  arm  in  each  lever  being 
12  in.  long.    What  weight  can  be  lifted  with  a  force  of  10  Ibs.? 

22.  If  the  short  arm  of  a  lathe  test  indicator  is  1  in.  long, 
and  the  length  of  the  pointer  over  all  is  14  in.,  what  is  the 


20  SHOP  MATHEMATICS 

amount  of  motion  at  the  end  when  indicated  point  is  T^Vv  m- 
from  axis  of  the  lathe  spindle? 

23.  A  windlass  is  to  raise  a  weight  of  1000  Ibs.  with  axle 
3  in.  dia.     What  length  of  crank  is  required  when  a  force 
of  120  Ibs.  is  to  be  used  on  end? 

24.  A  windlass  is  to  raise  1200  Ibs.  with  a  24  in.  crank. 
If  a  force  of  120  Ibs.  is  applied  at  the  end  of  crank,  what  is 
the  dia.  of  the  axle  required? 

25.  If  a  wheel  and  axle,  24  in.  dia.  of  wheel  and  3  in.  dia. 
of  axle,  required  25  Ibs.  force  to  raise  a  weight  of  200  Ibs., 
what  size  wheel  will  be  required  to  raise  500  Ibs.  with  an 
axle  2  in.  dia.  and  a  force  of  36  Ibs.? 

26.  A  windlass  with  an  18  in.  crank  and  3  in.  axle  will 
require  how  much  force  to  raise  a  weight  of  1500  Ibs.? 

27.  What  dia.  of  axle  will  be  required  for  a  windlass  with 
an  18  in.  crank  to  raise  a  weight  of  1000  Ibs.  with  a  force  of 
35  Ibs.? 

28.  What  dia.  of  axle  will  be  required  for  a  windlass  to 
raise  1600  Ibs.  with  a  20  in.  crank  and  a  force  of  45  Ibs.? 

29.  With  a  wheel  30  in.  dia.  and  axle  5  in.  dia.,  what 
force  will  be  required  to  raise  a  weight  of  500  Ibs.? 

30.  In  a  combination  of  three  wheels  and  axles,  when 
R  is  10  in.  for  each  wheel  and  r  is  4  in.  for  each  axle,  what 
weight  at  W  can  be  supported  by  a  pull  at  P  of  25  Ibs.? 

31.  If  P  moves  a  distance  of  5  ft.  in  problem  30,  how 
far  will  W  move  in  the  same  time? 

32.  Find  the  ratio  of  speeds  between  P  and  W  in  prob- 
lem 31. 

33.  How  many  R.  P.  M.  will  a  grinder  spindle  make  that 
is  driven  through  two  counter  shafts,  Fig.  13,  main  line  pulley 


21 

30  in.  dia.  and  150  R.  P.  M.  with  drivers  on  counters  each 
24  in.  dia.,  followers  6  in.  dia.  and  spindle  pulley  3  in.  dia.? 

34.  If  it  is  desired  to  drive  the  spindle  of  problem  33 
20,000  R.  P.  M.,  what  size  drive  pulley  will  be  required  on 
main  line  if  all  the  other  pulleys  remain  the  same? 

35.  A  spindle  is  driven  by  a  train  of  gears,  as  in  Fig.  12; 
the  first  driver  on  shaft  has  50  teeth,  the  two  stud  driving 
gears  each  50  teeth,  and  the  followers  each  125  teeth  with 
150  teeth  in  gear  on  spindle.    What  is  the  number  of  revolu- 
tions of  spindle  with  one  turn  of  the  shaft? 

36.  If  one  set  of  stud  gears  were  thrown  out  of  mesh  in 
the  train  of  problem  35  and  the  remaining  gears  moved  into 
mesh,  what  is  the  number  of  revolutions  of  spindle  for  one 
turn  of  the  shaft? 

37.  If  60  teeth  gears  were  substituted  in  the  place  of  the 
50  teeth  in  problem  35,  what  would  be  the  number  of  revo- 
lutions of  the  spindle  for  one  turn  of  the  shaft? 

38.  Find  the  "gear"  of  a  bicycle,  Fig.  14,  having  a  rear 
wheel  28  in.  dia.  with  18  teeth  in  front  and  7  teeth  in  rear 
sprocket. 

Note.  The  term  gear  as  used  in  problem  38  means  that  the  for- 
ward motion  due  to  one  revolution  of  the  crank  is  the  same  as  would 
be  produced  by  one  revolution  of  a  wheel  whose  diameter  in  inches 
is  equal  to  the  "gear." 

39.  What  is  the  "gear"  with  17  teeth  in  front,  8  teeth  in 
rear  sprocket,  with  30  in.  dia.  of  wheel? 

40.  What  is  the  "gear"  with  30  teeth  front,  10  teeth  rear 
and  28  in.  dia.  of  wheel? 

41.  W'hat  is  the  "gear"  with  16  teeth  front,  7  teeth  rear 
with  30  in.  dia.  of  wheel? 

42.  Find  the  "gear"  with  50  teeth  front,  15  teeth  rear 
and  28  in.  wheel. 


22  SHOP  MATHEMATICS 

43.  Find  the  "gear"  with  22  teeth  front,  7  teeth  rear  and 
26  in.  wheel. 

44.  What  is  the  "gear"  with  32  teeth  front,  12  teeth  rear 
and  26  in.  wheel? 

45.  What  is  the  "gear"  with  24  teeth  front  and  7  teeth 
rear  sprocket  and  24  in.  dia.  of  wheel? 

46.  A  man  weighing  150  Ibs.  has  to  raise  a  weight  of  1,200 
Ibs.    How  many  sheaves  must  be  placed  in  each  pulley  block 
as  arranged  in  Fig.  17  to  raise  the  weight? 

47.  What  force  would  be  required  in  problem  46  if  a 
pulley  is  used  as  in  Fig.  19,  when  sheave  B  is  8  in.  dia.  and  C 
is  7  in.  diameter? 

48.  With  the  pulley  blocks  as  in  Fig.  17,  with  4  sheaves 
in  both  A  and  B,  what  weight  can  be  raised  with  a  15  Ib. 
pull  at  P? 

49.  What  weight  can  be  lifted  with  a  pair  of  6  sheave 
pulley  blocks,  as  in  Fig.  17,  with  an  85  Ib.  pull  at  P? 

50.  When  the  pull  of  85  Ibs.  is  in  the  direction  shown  in 
Fig.   18,  what  weight  can  be  raised  with  6  sheave  pulley 
blocks? 

51.  What  is  the  ratio  of  the  efficiency  of  the  pulley  blocks 
of  problems  49  and  50? 

52.  What  force  will  be  required  to  lift  a  1,200  Ibs.  ma- 
chine with  the  pulley  blocks  of  problem  48? 

53.  How  many  men  weighing  150  Ibs.  each  can  raise,  by 
their  own  weight,  a  block  of  stone  that  weighs  10,000  Ibs. 
with  a  pulley  block  as  in  Fig.  19,  when  sheaves  B  and  C  are 
10  in.  and  9  in.  dia.  respectively? 

54.  Find  the  size  of  a  differential  pulley,  to  raise  6  tons 
with  a  force  of  150  Ibs.  when  the  difference  of  the  diameters 
is     inch. 


SHOP  MATHEMATICS  23 

IV.     INCLINED  PLANE 

The  inclined  plane  is  a  flat  surface  sloping  or  inclined  to 
the  horizontal.  With  the  inclined  plane  a  weight  can  be 
raised  by  a  force  of  less  magnitude  as  the  following  illus- 
tration will  show. 

Illustration.  Suppose  the  weight  W  is  to  be  raised  from 
a  horizontal  AC  to  a  point  D, 
Fig.  21.  If  the  weight  is  raised 
in  a  vertical  line,  as  CD,  then  P 
and  W  act  through  the  same  dis- 
tance .'.  by  the  general  law  of 
machines  P  =  W.  But  if  the 
weight  is  pushed  up  the  incline 
AD,  then  the  force  acts  through 
the  distance  AD  while  the  weight  is  lifted  only  the  distance 
CD;  and  the  statement  becomes 

P  :W  =  CD  :AD 

or  P  :  W  =  height  of  plane  :  length  of  plane, 

from  which  are  obtained  the  formulas 
WXH 


P- 

and  W  = 


L 

PXL 


H 

If  the  force  acts  along  a  line  parallel  to  the  base,  AC 
then 

P  :  W  =  height  :  base 

from  which  is  obtained 
WXH 
B 

w     PxB 
and       fF=_- 


24  SHOP  MATHEMATICS 

If  the  force  acts  at  any  angle  to  the  plane  as  Y  Fig.  21 A 
then  P  \W  =  sin  x  :  cos  Y 

WXsinx  PXcos  Y 

and  P= ^—  W  = : 

cos  Y  sin  x 

Example.  A  horse  pulling  a  load  on  the  level  has  only  the 
friction  to  overcome,  but  the  moment 
that  it  starts  up  an  incline  it  has  a 
part  of  the  weight  of  the  load  added 
to  the  force  required  to  overcome  fric- 
tion. If  the  pull  was  500  lbs.  on  the 
level  and  the  load  weighs  1,200  lbs., 
what  extra  force  is  required  on  an 
incline  of  1  foot  in  20  feet? 

Solution.     By  formula: 

P  = — f —  =  — — '=60.     Then  60+500  =  560  lbs. 

L  20 

V.     THE  SCREW 

The  screw  is  an  inclined  plane  wrapped  or  wound  around 
a  cylinder.  If  the  incline  is  long  in  comparison  with  the 
diameter  of  the  cylinder,  it  may  extend  more  than  once 
around  the  cylinder  forming  the  threads  of  the  screw;  the 
height  of  the  incline  in  going  once  around  is  called  the  lead 
of  the  screw.  The  term  pitch  is  used  to  designate  either  the 
number  of  threads  per  inch  or  the  distance  from  the  top  of  one 
thread  to  the  top  of  the  next;  hence  in  a  single  thread  screw 
the  lead  is  the  same  as  the  pitch,  but  in  a  double  or  triple 
thread  screw  the  lead  is  two  or  three  times  the  pitch. 

When  the  screw  is  turned  on  its  axis  through  one  revolu- 
tion, the  nut  being  stationary,  the  screw  is  raised  or  lowered 
a  distance  equal  to  the  lead  of  the  thread;  the  force  moves 
in  the  same  time  a  distance  equal  to  the  circumference  of 
twice  the  length  of  the  lever  or  bar  that  is  used  to  turn  the 


SHOP  MATHEMATICS 


25 


screw  on  its  axis.    From  this  is  obtained  the  following: 

RULE.    The  force  multiplied  by  the  circumference  of  the 

circle  through  which  the  force  arm  moves  equals  the  weight 

multiplied  by  the  lead  of  the  screw. 

From  this  rule  is  obtained  the  statement: 

P  :W=L  :2irR 


shown  in  Fig.  23,  the 
differential  screw.  This 
is  made  with  two  screws 
of  different  pitches,  or 
leads  of  threads,  either 
both  right  or  both  left 
hand  threads. 

It  will  be  seen  from 
Fig.  23  that  one  turn  of 
the  large  screw  lifts  the 
weight  only  the  differ- 
ence between  the  leads 
of  the  large  and  small 
screws;  then  by  the  Gen- 
eral Law. 


where 

wrench 
screw 

then, 


L  =  lead  of  screw 
R  =  length  of  bar  or 
used   to   operate  the 


P  = 


WXL 


L 

The  screw  can  be  com- 
pounded like  the  other  ele- 
ments of  machines,  as 


Note.     The  nut  N,  Fig.  22,  is  the  part  that  must  be  used  with  a 
screw  to  make  it  effective. 


26  SHOP  MATHEMATICS 


and 


P  :  W  =  L—l  :2 
WX(L—l) 


L—l 

Note.  Owing  to  the  amount  of  friction  in  the  differential  screw 
its  practical  use  is  limited. 

VI.     WEDGE 

The  wedge  is  a  pair  of  inclined  planes  placed  back  to  back. 
It  is  used  in  two  ways;  by  being  driven  with  a  blow  of  the 
hammer,  and  by  pressure  which  usually  acts  parallel  to  the 
base  of  the  planes.  The  difficulty  in  calculating  the  effective- 
ness of  the  first  kind  is  to  determine  the  force  of  the  hammer 
blow,  otherwise,  the  statement  and  rule  is  the  same  for  either 
kind. 

.-.P  :  W=T  :L 

WXT 


or 


L 

PXL 


T 
where      T=  thickness. 

The  rule  is  thus  the  same  as  for  the  inclined  plane  where 
force  acts  parallel  to  the  base  of  incline. 

Example.       A 
sliding  wedge, 
Fig.    24,    has    a 
pressure   of   100        Jfyq  24 
Ibs.,  the  width  of 
back  end  is  4  in.,  the  length  is  20  in.     What  weight  can 
be  raised  by  its  use? 


SHOP  MATHEMATICS  27 


Solution.     By  formula: 

TT7     PXL     100X20 

W  =  — =—  = =  500  Ibs. 

l  4 


MISCELLANEOUS  PROBLEMS 

Note.     In  the  following  problems  friction  of  moving  parts  will  not 
be  considered. 

1.  An  iron  ball  weighing  398  Ibs.  rests  on  a  surface  which 
is  inclined  16°  45'  to  the  horizontal.    What  force,  acting  at 
an  angle  of  14°  30'  to  the  incline,  is  required  to  hold  the  ball 
in  position? 

2.  A  weight  of  3,500  Ibs.  is  to  be  drawn  up  an  incline 
640  ft.  long,  85  ft.  above  the  horizontal.    What  force  acting 
parallel  to  incline  will  be  required  to  keep  the  load  on  incline? 

3.  A  cylinder  of  cast  iron  24  in.  dia.,  30  in.  long  is  to  be 
rolled  up  an  incline  of  18°  15'.    What  force,  acting  at  8°  15' 
to  the  incline,  will  be  required  to  hold  the  cylinder  from 
rolling  down? 

4.  What  weight  will  be  raised  with  a  screw  of  £  in.  lead 
when  100  Ibs.  of  force  is  applied  at  the  end  of  a  lever  18  in. 
from  the  center  of  screw? 

5.  When  the  lead  of  a  screw  is  f  in.,  R  is  20  in.  and  a 
weight  of  12,000  Ibs.  is  to  be  raised,  what  is  the  force  required? 

6.  In  Fig.  22,  a  screw  of  ^  in.  lead  is  turned  with  a 
bar  1^  in.  long,  with  1  Ib.  of  force  on  end  of  R;  what  weight 
can  be  raised? 

7.  A  sliding  wedge,  Fig.  24,  is  used  to  raise  the  knife 
on  a  shearing  press  that  weighs  100  Ibs.,  the  wedge  is  to  move 
18  in.  and  is  3  in.  thick  at  back  end.    What  force  will  be  re- 
quired? 

8.  A  truck  loaded  with  an  engine  weighing  6  tons  is  to 
be  drawn  up  an  incline  12  ft.  long  and  5  ft.  above  the  hori- 


28  SHOP  MATHEMATICS 

zontal.    What  force  will  be  required  when  the  pull  is  parallel 
to  the  incline? 

9.     What  weight  can  be  drawn  up  an  incline  10  ft.  long 
and  4  ft.  high  with  a  pull  of  300  Ibs.? 

10.  Two  men  each  pulling  125  Ibs.  can  pull  what  weight 
up  an  incline  8  ft.  long  and  6  ft.  high? 

11.  What  force  will  be  required  on  a  single  thread  screw 
having  3  threads  per  inch,  with  a  bar  18  in.  long  from  center 
of  screw  to  raise  a  block  of  granite  that  weighs  5  tons? 

12.  Two  jack  screws  are  to  be  used  to  raise  a  block  10  ft. 
long,  weighing  10,000  Ibs.     One  is  a  third  more  powerful 
than  the  other.    Make  sketch  showing  the  position  of  screws 
to  give  the  proportionate  load  on  each. 

13.  How  many  jack  screws  with  \  in.  lead  and  having 
ft  16  in.  long,  will  be  required  to  raise  a  building  weighing 
50  tons,  if  the  pull  on  each  lever  is  50  Ibs.  ? 

14.  How  many  jacks  will  be  required  with  screws  of  £  in. 
lead,  and  12  in.  levers,  with  25  Ibs.  pull  on  each  lever  to  raise 
the  building  of  problem  13? 

15.  A  wedge  8  in.  long,  If  in.  thick  at  end  will  require 
how  many  Ibs.  of  a  hammer  blow  to  drive  it  into  a  log  that 
has  a  resistance  of  2,000  Ibs.  against  splitting? 

16.  When  the  screws  of  a  differential  are  8  and  12  pitch 
single  threads  respectively,  with 

a  pull  of  5  Ibs.,  what  length  of 
lever  will  be  required  to  raise  a 
weight  of  5, 000  Ibs.? 

17.  What  length  of  bar  will 
be  required  to  raise  a  building  of 
100  tons  weight  with  10  Ibs.  pull 
each  on  100,  \  in.  lead  jack  screws? 

18.  A    cylinder    25    in.    dia. 
weighs  5,000  Ibs.,  Fig.  26.     What 

force  at  P  will  hold  the  cvlinder  on  incline? 


SHOP  MATHEMATICS 


29 


19.  If  the  cylinder  of  problem  18  weighs  6,575  Ibs.,  the 
lever  is  36?  in.  from  P  to  fulcrum,  12|  in.  from  weight  to 
incline,  and  incline  is  10  ft.  long,  4  ft.  high,  what  force  on 
the  end  of  lever  P  will  prevent  cylinder  from  rolling  back? 

SCREW  THREADS 

The  formula  for  finding  tap  drill  sizes  is  based  on  the  depth 
from  point  to  bottom  of  thread, 
as  follows:  Fig.  27  shows  outline 
of  the  sharp,  or  V  thread.  The 
depth  A  of  a  thread  of  1  inch  pitch  is 
found  by  trigonometry  and  is  equal 
to  .8660  inch.  If  then  the  thread  is 
taken  on  both  sides  of  a  cylinder 
the  double  depth  for  1  inch  pitch  =  .8660X2  =  1. 732. 

1  732 
For  any  other  pitch  of  thread  the  double  depth  =    ' 

where    N  is  the  number  of  the  threads  per  inch  of  the 
required  screw. 

Then  the  formula  for  tap  drill  size  is 
1.732 


Fiy.27. 


S=T— 


N 


S  =  diameter  of  drill. 

T  =  outside  diameter  of  bolt. 

Fig.  28  shows  the  outline  of  the  United  States  Standard 
(U.  S.  S.)  thread.  Here  one- 
eighth  of  the  total  depth  of  the 
sharp  V  is  flattened  on  the 
points,  and  the  same  amount 
filled  in  at  the  bottom  of  the  V, 
thus  making  the  flattened  parts 
of  the  U.  S.  S.  thread  each  equal 
to  one-eigth  of  the  pitch  so  that  the  double  depth  for  a 


2^.28. 


30  SHOP  MATHEMATICS 

U.  S.  S.  thread  of  1  inch  lead  will  be 

1.732— 2 X\  of  1.732 
or  1.732— .432=1.3 

1  3 

For  any  other  pitch  the  double  depth  will  be  -^=- 

where     T  =  outside  diameter  of  bolt. 

Then  S  is  the  size  of  tap  drill  to  be  used  before  tapping 

1  3 

threads  in  the  nut.    The  formula  is  S=  T ~ . 

jN 

Example.     For  a  ^X  13  pi.  U.  S.  S.  bolt,  what  is  the  size 
of  tap  drill? 

Solution.     By  formula : 

s=  T  -  '-=    ~    -    -    -  •**> inch  ta drm- 


It  is  sometimes  necessary  to  use  the  British  Standard 
thread,  or  Whitworth  Standard,  as  it  is  usually  called,  after 
the  name  of  the  man  who  established  it. 

In  this  thread  the  top  is 
rounded  over  one-sixth  of  the 
depth  and  the  bottom  is  filled  in 
the  same  amount,  so  that  one- 
third  of  the  depth  is  subtracted 
for  actual  depth  of  thread, Fig.  29. 

I  28 
The  formula  for  tap  drill  is,    S  =  T ^-. 

The  included  angle  of  the  thread  is  55°  instead  of  60C 
as  in  the  V  and  U.  S.  S.  threads. 

PROBLEMS 

1.  Find  the  depth  of  a  V  thread  of  12  pitch. 

2.  Find  the  depth  of  an  18  pi.  V  thread. 

3.  What  is  the  depth  of  a  U.  S.  S.  thread  of  12  pitch? 


SHOP  MATHEMATICS  31 

4.  What  is  the  double  depth  for  a  Whitworth  thread  of 
11  pitch? 

5.  What  is  the  tap  drill  size  for  a  1  in.  by  8  pi.  U.  S.  S. 
tap? 

6.  What  is  the  tap  drill  size  for  ^  in.  by  20  pi.  V  thread 
nut? 

7.  What  size  drill  will  be  needed  to  allow  a  full  thread  on 
a  £  in.  by  12  pi.  Whitworth  tap? 

8.  Give  the  width  of  the  flat  on  the  top  of  a  £  in.  by  9  pi. 
U.  S.  S.  thread. 

9.  Find  tap  drill  size  for  18  pi.  double  V  thread  nut 
^  in.  diameter. 

10.  What  is  the  bottom  dia.  of  a  f  in.  by  11  pi.  U.  S.  S. 
thread? 

11.  Find  bottom  dia.  of  a  f  in.  by  10  pi.  Whitworth 
thread. 

12.  Find  bottom  dia.  of  a  £  in.  by  13  pi.  V  thread. 

13.  What  is  the  double  depth  for  a  f  in.  by  16  pi.  U.  S.  S. 
bolt? 

14.  What  is  the  double  depth  for  a  \  in.  by  20  pi.  V  thread? 

15.  If  there  are  16  threads  per  inch  on  a  1  in.  dia.  bolt 
and  the  nut  advances  £  in.  for  each  turn  of  the  bolt,  what  is 
the  pitch  of  the  thread? 

16.  One  turn  of  the  feed   screw  on  a  milling  machine 
moves  the  table  ^  in.,  but  the  threads  measure  8  per  in. 
How  should  the  threads  be  designated? 

Answer.     %  in.  lead  or  8  pitch  double  thread. 

17.  If  there  were  12  threads  per  in.,  and  one  turn  of  screw 
moves  nut  i  in.,  how  would  the  thread  be  designated? 


32 


SHOP  MATHEMATICS 


TOOTHED  GEAR  WHEELS 


A  toothed  gear  wheel  is  one  with  projections  on  its  periphery  ; 
these  projections,  or  teeth,  engage  with  the  teeth  of  a  similar 
wheel  and  the  engagement  or  meshing  of  these  teeth  imparts 
a  positive  rotary  motion  from  driver  to  follower. 

The  ratio  of  the  speeds  of  two  gears  that  run  together  is 
called  their  velocity  ratio,  and  is  in  inverse  proportion  to  their 
size. 

Of  two  gears,  if  one  revolves  once  while  the  other  revolves 
twice,  their  velocity  ratio  is  as  1  to  2=^,  which  indicates 
that  the  first  gear  is  twice  the  size  of  the  other. 

The  pitch  circle  of  a  gear  is  the  circle  near  the  center  be- 
tween top  and  bottom  on  the  face  of  the  teeth,  such  that  if 
the  teeth  were  to  be  made  infinitely  small  the  gear  wheel 
would  become  a  cylinder,  Fig.  31. 


Fly.  31. 


The  diameter  of  a  gear  wheel  is  always  the  diameter  of  the 
pitch  circle,  unless  otherwise  stated. 

The  outside  diameter  is  the  diameter  of  the  blank  in  which 
the  teeth  of  a  gear  are  cut. 

The  pitch  of  the  teeth  of  a  gear,  usually  called  the  diametral 
pitch,  is  the  number  of  teeth  for  each  inch  of  the  diameter  of  the 
gear,  unless  the  circular  pitch  is  stated,  which  is  the  distance 


SHOP  MATHEMATICS  33 

from  center  of  one  tooth  to  center  of  the  next  tooth  measured  on 
the  pitch  circle. 

When  a  pair  of  gears  are  running  in  mesh,  the  smaller  of 
the  pair  is  called  the  pinion. 
Let     D    =  diameter    of   pitch    circle    of   gear,   and   OD  = 

diameter  of  blank. 
P    =  diametral  pitch. 
N    =  number  of  teeth  in  large  gear  and  n  =  number  of 

teeth  in  pinion. 

C    =  circumference  of  pitch  circle. 
CP=  circular  pitch. 

T    =  thickness  of  tooth  on  pitch  circle, 
a     ==  addendum. 
x     =  distance  between  centers  of  two    gears    in   mesh 

with  each  other. 
\r      \r  4-  &      TT 

TVimi     P 

D       OD~     CP 

N  2 

D    — p-02>— p 


C     = 


CP      1.57 


„.  T      .151 

Clearance 


— 
10        P 


34  SHOP  MATHEMATICS 

<2? 

Working  depth   = 


Whole  depth       = 


P 

2.157 


_ 

o 

The  thickness  of  cast  gears  with  cut  teeth  =  -^ 

To  find  the  diameters  of  two  gears  in  mesh  with  each  other 
when  x  and  velocity  ratio  of  the  two  gears  are  given: 

RULE.  Divide  the  distance  between  the  centers  by  the 
sum  of  the  terms  of  the  ratio;  the  pitch  diameters  will  be 
twice  the  quotient  multiplied  by  each  term  of  the  ratio. 

Example.  Find  the  diameter  of  two  gears  with  centers 
10  in.  apart  and  a  velocity  ratio  of  2  to  5. 

Solution.     By  rule: 


10  +  7  =  142857 

1.42857X2X2  =  5.72 
and        1  42857  X2X5  =  14.28 

5.72  in.  and  14-28  in.  are  the  pitch  diameters  of  gears. 

PROBLEMS 

1.  What  is  the  circular  pitch  of  an  8  pi.  gear  of  40  teeth? 

2.  What  is  the  circular  pitch  of  a  6  pi.  gear  of  108  teeth? 

3.  What  is  the  pitch  of  a  75  tooth  gear  when  the  circular 
pitch  is  .7854? 

4.  A  gear  of  45  teeth  has  a  circular  pitch  of  .3142  in. 
Find  the  diametral  pitch  of  gear. 

5.  What  is  the  dia.  of  a  gear  blank  that  has  6  pi.  teeth 
and  18  in.  diameter? 


SHOP  MATHEMATICS  35 

6.  A  gear  blank  is  3|  in.  dia.  and  is  to  have  13  teeth. 
What  is  the  pitch? 

7.  What  is  the  number  of  teeth  in  a  6  pi.  gear  18  in. 
diameter? 

8.  What  is  the  number  of  teeth  in  a  4  pi.  gear  19^  in. 
blank  diameter? 

9.  WThat  is  the  thickness  of  tooth  for  a  75  tooth  gear, 
.7854  in.  circular  pitch? 

10.  What  is  the  thickness  of  tooth  for  a  4  pi.  gear  of  40 
teeth? 

11.  Find  the  whole  depth  of  tooth  for  a  4  pi.   gear  of 
40  teeth. 

Xote.     Unless  otherwise  stated  the  depth  of  tooth  means  the 
whole  depth. 

12.  What  is  the  depth  of  tooth  for  a  6  pi.  gear  with  30 
teeth? 

13.  Find  dia.  of  blank  for  a  16  pi.  gear  with  32  teeth. 

14.  What  is  the  dia.  of  blank  of  a  10  pi.  gear  with  28  teeth? 

15.  What  is  the  pitch  of  a  75  tooth  gear  19^  in.  outside 
diameter? 

16.  Find  the  number  of  teeth  in  a  16   pi.   gear  8   in. 
diameter. 

17.  Find  the  number  of  teeth  in  a  10  pi.  gear  64  in. 
diameter. 

18.  Find  the  number  of  teeth  in  a  20  pi.  gear  10.1  in. 
outside  diameter. 

19.  Find  the  sizes  of  two  gears  when  the  velocity  ratio  is 
1  to  2  and  the  distance  between  centers  is  12  inches. 

20.  Find  the  diameters  of  two  gears  12  in.  between  cen- 
ters, with  velocity  ratio  2  to  4. 

21.  Find  the  diameters  of  two  gears  28  in.  between  cen- 
ters, when  the  velocity  ratio  is  3  to  5. 


36 


SHOP  MATHEMATICS 


22.  When  the  distance  between  centers  of  two  gears  is 
27  in.  and  the  velocity  ratio  is  4  to  5,  find  the  diameters  of 
the  gears. 

23.  When  velocity  ratio  of  two  gears  is  4  to  5,  and  the 
distance  between  the  centers  is  36  in.,  find  the  diameters  of 
the  gears. 

24.  The  velocity  ratio  of  two  gears  is  2  to  3  and  distance 
between  the  centers  30  in.    Find  the  diameters  of  gears. 

25.  What  is  the  clearance  of  a  4  pi.  gear  of  75  teeth? 

26.  Find  the  clearance  of  teeth  for  a  2  in.  dia.  gear  with 
32  teeth. 

27.  Find  the  dia.  of  a  6  pi.  gear  with  108  teeth. 

28.  What  is  the  circumference  of  a  gear  blank  with  32 
teeth  16  pitch? 

29.  Two  gears  in  mesh  have  30  and  24  teeth  respectively 
and  are  6  pitch.    W^hat  is  the  distance  between  centers? 

30.  Two  gears  running  together  have  120  and  80  teeth 
respectively  and  are  10  pitch.     Find  the  distance  between 
the  centers. 

Emery  wheels  are  designed  to  run 
at  5,500  F.  P.  M.  (surface  feet  per 
minute) ,  but  the  speeds  of  wheels  are 
usually  given  in  R.  P.  M.; 

F.  P.  M.  =  R.  P.  M.   X   irD, 
D  =  diameter  of  wheel  in  feet. 
F.  P.  M. 


R.  P.  M.= 


rD 


31.  A  hand  power  grinding  wheel,  W,  Fig.  32,  is  6  in.  dia. 
and  is  to  run  5,500  F.  P.  M.  What  is  the  R.  P.  M.  of  crank 
C  when  A  has  200  teeth,  B  20  teeth,  D  15  teeth,  and  E  250 
teeth? 


SHOP  MATHEMATICS  37 

32.  How  many  R.  P.  M.  of  W.  in  problem  31,  when 
crank  makes  25  R.  P.  Af .? 

33.  How  many  F.  P.  M.  for  x  in  problems  31  and  32 
when  C  is  12  in.  long? 

34.  If  there  is  a  resistance  of  3  Ibs.  on  wheel  W,  problem 
31,  what  force  will  be  required  at  x  to  turn  wheel  5,000 
F.  P.  MJ 

35.  What  is  the  velocity  ratio  between  the  R.  P.  M.  of 
W  and  x  in  problem  31? 

Polishing  wheels,  with  wood  center  and  leather  covered 
rim,  are  allowed  to  run  7,000  F.  P.  M. 

Cloth  buffing  wheels  also  run  7,000  F.  P.  M. 

Hair-brush  cleaning  wheels  run  12,000  F.  P.  M. 

Ordinary  grindstones  (Ohio)  should  not  run  over  2,500 
F.  P.  M.  and  Huron  stones  3,500  F.  P.  M. 

36.  If  a  Huron  stone  of  same  size  as  in  problem  31  was 
substituted  for  the  emery  wheel,  what  R.  P.  M.  of  C  will  give 
3,500  F.  P.  M.  of  W? 

37.  If  a  brush  wheel  4^  in.  dia.  was  substituted  for  wheel 
of  problem  31  to  run  10,000  F.  P.  M.,  find  R.  P.  M.  of  C. 

38.  What  force  will  be  required  at  x  if  there  is  1  Ib. 
resistance  at  surface  of  W  in  problem  37? 

39.  What  is  the  R.  P.  M.  of  C  in  problem  31  when  A, 
B,  D,  and  E  are  changed  to  friction  wheels  9  in.,  1\  in.,  2  in., 
and  11  in.  diameters  respectively? 

BEVEL  GEARS 

Positive  rotary  motion  can  be  transmitted  from  driver 
to  follower,  when  the  shafts  stand  at  an  angle  to  each  other, 
and  in  the  same  plane,  by  using  a  pair  of  bevel  gears. 


38 


SHOP  MATHEMATICS 


When  the  angle  of  the  shafts  is  90°  and  the  velocity  ratio 
is  1  to  1  ,  both  gears  are  of  the  same  size  and  are  called  miter 
gears.  When  the  pair  run  at  a  velocity  ratio  other  than 
1  to  1,  the  smaller  gear  is  called  the  pinion,  the  same  as  in 
spur  gears. 

The  pitch  diameter  of  a  bevel  gear  is  the  diameter  of  the 
base  of  the  pitch  cone,  AB.  All  calculations  for  sizes,  as  in 
spur  gears,  are  made  on  this  base  circle.  The  teeth  of  bevel 
gears  vanish  or  become  infinitely  small  at  the  apex  of  the 
pitch  cone.  O^AM  is  the  normal  cone,  that  is,  one  whose  con- 
vex surface  stands  at  right  angles  to  the  pitch  cone  0AM. 
The  shape  and  size  of  teeth  are  found  on  the  normal  cone. 


The  addendum 


=  -    and  P,  D,  N,  C,  CP,  and  T  also 


clearance,  whole  and  working  depths  of  teeth  are  all  found  by 
the  formulas  on  page  33  for  spur  gears. 

The  calculations  for  sizes  and  shapes  of  blanks  are  found 
by  the  following  formulas: 


HK-H 


D  =  diameter  of  gear. 
Dj  =  diameter  of  pinion. 
C  =  center  angle  of  gear. 
Cl  =  center  angle  of  pinion, 


SHOP  MATHEMATICS  39 


AF 
then,        tan  C  =  --=,  by  (5)  page  179 


and         Cl  =  90-C  or  tan  C,  =  -^ 

(Jjb 

~~P 

and        ASl  =  a+  clearance. 

Angle  FOS=£C+£AOS 

An*     AS 

and         tan  AOb  =  —r-^ 
AO 

Angle  AOS  is  called  the  angle  of  increment,  angle 
AOSi  is  the  angle  of  decrement  for  the  gear. 

AS, 
tan  AOS^—rj^.     This  angle  is  used  in  the  setting 

of  the  blank  at  the  proper  angle  for  the  depth  of  tooth. 

The  OD  for  the  bevel  gear = D  +2H 
and        H  =  cos  CxAS  (by  4)  page  179. 

Example.  Two  shafts  at  an  angle  of  90°  to  each  other  are 
connected  by  a  pair  of  16  pitch  bevel  gears  with  a  velocity 
ratio  of  3  to  4.  If  the  pinion  has  48  teeth  find  all  the  other 
dimensions  of  the  gears. 

Solution.  If  the  pinion  has  48  teeth  and  the  velocity 
ratio  is  3  to  4  then  3  :4  =  4$  '  X  which  makes  X=64,  the 
number  of  teeth  in  the  gear. 

D=—  =  4  inches.     D.  =  —  =  3  inches. 
lo  16 

AF      2 

tn<n     f\f     /  C1 1    QQQQ    *      /  C* KQO    O' 

lLift/       \J  I      ./      \_s    ~^T    ^TT  —  J_  .OOOO    •      •    /      \J    t/O         O      • 

j  ^^     1.5 


=  — =  1-  =  . 0625;  ASl=  —  +—=. 0723. 
p      16  P        P 


40  SHOP  MATHEMATICS 


tan  of 

&. 

Face  angle  of  gear  •=  Z  FOS=  Z  C  +  Z  AOS  =  54°  34'. 


AU        2.5 
Z  FOS,  =  Z  C—  Z  AOS,  =  51°  29'. 

This  is  the  angle  of  working  depth,  and  is  the  angle  at 
which  to  set  the  spiral  head  for  cutting  the  teeth  of  the  gear. 
OD  of  gear  blank=D+2H. 

SIT 

H  =  ST;cos  ^AST=~=cos  ZCV.  #  =  ASXcos  ZC 
/lo 

then  OD  =  4+2X  .031  '5  =  4.  07  '5  inches. 
The  dimensions  of  the  pinion  can  be  found  in  a  similar 
manner. 

PROBLEMS 

1.  Two  bevel  gears,  shafts  at  90°,  have  a  velocity  ratio 
of  2  to  3  with  40  6  pi.  teeth  in  the  pinion.    What  are  the 
diameters  and  angles  of  blanks? 

2.  In  a  pair  of  bevel  gears  with  shafts  at  90°,  having  a 
velocity  ratio  4  to  5,  the  gear  is  15  in.  dia.  and  has  8  pi.  teeth. 
Find  diameters  and  angles  to  cut  teeth  of  gear  and  pinion. 

3.  The  circular  pitch  of  a  miter  gear  is  .31416  in.  and 
it  has  40  teeth.    Find  sizes  and  angles  required  to  machine 
the  gear. 

4.  The  thickness  of  the  tooth  of  a  bevel  gear  pinion  of 
32  teeth  is  .19635  inches.     What  are  the  finished  sizes  and 
angles  of  a  pair  of  gears  with  velocity  ratio  of  3  to  5,  when 
shafts  stand  at  90°  to  each  other? 


SHOP  MATHEMATICS  41 

5.  The  center  lines  of  a  pair  of  bevel  gears  of  6  pitch 
stand  at  90°  to  each  other  and  the  velocity  ratio  is  3  to  4. 
Find  finished  sizes  and  angles  of  gears  when  pinion  has  30 
teeth. 

6.  Two  bevel  gears  with  shafts  at  right  angles  to  each 
other  have  a  velocity  ratio  of  4  to  5;  the  pinion  is  15  in.  dia. 
with  8  pitch  teeth.    What  are  the  diameters  and  angles  of 
the  pair? 

WORM  GEARING 

When  two  shafts  stand  at  90°  to  each  other  but  not  in  the 
same  plane,  rotary  motion  can  be  transmitted  between  the 
shafts  by  the  use  of  worm  gearing.  This  is  generally  used 
where  a  great  difference  is  required  in  the  velocity  ratio, 
combined  with  positive  action  between  the  teeth. 

There  is  one  great  objection  to  worm  gearing  on  account  of 
the  friction  due  to  the  sliding  action  between  the  teeth.  The 
worm  is  a  screw  with  threads  shaped  on  the  plan  of  the  U.  S. 
S.  except  that  the  flats  at  the  top  are  much  wider  and  the 
included  angle  of  the  sides  is  29°.  The  thread  of  a  worm 
can  be  spoken  of  as  having  a  lead  and  it  can  be  of  single, 
double,  triple,  quadruple,  etc.,  thread  as  in  screw  threads. 

The  tooth  of  the  worm  wheel,  which  meshes  with  the  thread 
of  the  worm,  advances  the  distance  of  one  tooth  at  each 
revolution  of  a  single  thread  worm;  therefore  if  a  worm 
wheel  has  30  teeth,  the  worm  will  have  to  revolve  thirty 
times  for  one  complete  revolution  of  the  worm  wheel. 

The  velocity  ratio  in  the  case  of  double,  triple,  etc.,  thread 
worms  will  then  be  the  number  of  teeth  in  the  worm  wheel 
divided  by  1,  2,  3,  etc.,  according  as  it  is  a  single,  double, 
triple,  etc.,  thread  worm. 


42 


SHOP  MATHEMATICS 


PROBLEMS 

1.     A,  Fig.  34,  is  a  double  thread  worm  and  B  a  worm 

wheel  with  80  teeth, 
12 J  in.  dia.  fast  to  an 
axle,  a,  4  in.  dia.  The 
crank  C  is  12  in.  long. 
What  weight  can  be  bal- 
anced at  W  when  75  Ibs. 
of  force  is  applied  at  xf 

2.  If  A,  Fig.  34,  had 
been  a  triple  thread 
worm  and  B  had  45  teeth  and  18  in.  dia.  with  5  in.  axle,  a, 
what  length  of  crank  C  will  be  required  to  lift  4,000  Ibs. 
with  a  force  of  60  Ibs.  at  x? 

3.  If  a,  Fig.  34,  is  a  differential  axle  with  5  in.  and  7  in. 
diameters,  what  length  of  crank  C  will  be  required  to  lift 
24,000  Ibs.  at  W,  with  60  Ibs.  at  x,  when  B  is  16  in.  dia.  with 
45  teeth  and  A  is  a  single  thread  worm? 

4.  If  on  left  end  of  A,  Fig.  34,  a  gear  of  60  teeth  is  in 
mesh  with  a  gear  of  24  teeth  which  is  turned  by  an  18  in. 
crank  Cl}  when  A  is  single  thread  worm,  B  15  in.  dia.  with 
42  teeth,  a  4  in.  dia.,  what  weight  can  be  raised  with  a  50  Ib. 
pull  at  Y? 

5.  With  mechanism  the  same  as  in  problem  4  except  a 
differential  axle  at  a,  having  4^  in.  and  6£  in.  diameters, 
what  weight  can  be  raised? 

SPIRAL  GEARING 

Worm  gearing  is  the  limiting  case  of  what  is  called  spiral 
or  helical  gearing  with  the  velocity  ratio  expressed,  in  a  single 

thread  worm,  as  -^ -r- — -r— : =• — r    The  other  extreme  is 

No.  oj  teeth  in  wheel. 


SHOP  MATHEMATICS  43 

where  the  velocity  ratio  is  1  to  1,  with  the  shafts  in  different 
planes  and  at  any  angle  to  each  other  from  90°  to  0°.  When 
the  shafts  make  an  angle  of  90°  with  each  other,  the  angle  of 
spiral  for  each  wheel  will  be  45°,  the  spiral  being  either  right 
or  left  hand.  If,  however,  the  shafts  are  parallel,  the  teeth 
of  one  gear  will  be  right  hand  spiral  and  the  other  left  hand 
spiral  and  the  angle  of  spiral  needs  to  be  only  large  enough 
to  have  each  tooth  begin  its  contact  with  the  tooth  of  the 
mating  gear  before  the  next  tooth  ends  its  contact  with  the 
next  tooth  of  the  mating  gear. 

The  size  of  the  blank  is  found  in  a  different  manner  from 
that  of  the  spur  wheel  formulas  (although  cutters  of  the  same 
pitch  may  be  used  to  cut  the  teeth) ,  and  must  be  calculated 
as  follows: 

,  Let  Fig.  35  represent  the  top  view  of  a 

blank  for  a  helical  gear  with  lines  1  and 
2  drawn  through  the  centers  of  two 
adjacent  teeth  at  the  angle  which  the 
teeth  make  with  the  axis  of  the  gear. 
The  distance  Pl  measured  at  right  angles 
H p  r~  to  the  lines  1  and  2  is  called  the  normal 

fta.35.         circular   pitch  and  is  calculated    as   for 
ordinary   spur  wheels  according  to   the 
pitch  of  cutters  used  to  form  the  teeth.     The  distance  P 
on  the  face  of  the  gear,  called  the  circular  pitch  determines 
the  size  of   the   gear,  after  the  number  of  teeth  is  given, 


p 

then  by  trigonometry  P  =  — 


Example.     Two  helical  gears,  with  velocity  ratio  1  to  1 
shafts  at  90°  are  to  be  cut  with  a  Brown  &  Sharpe  16  pitch 
cutter.    Find  the  size  of  blanks  required  for  20  teeth. 


44  SHOP  MATHEMATICS 

Solution.  As  angle  of  shafts  is  90°  and  gears  are  the  same 
size,  angle  F  =  45°,  and  may  be  either  right  or  left  hand. 
The  normal  circular  pitch  Pl  for  16  pitch  cutter  =.19636, 
then, 

by  formula 

.19636 _ 
.70711 
Now  as  the  gears  are  to  have  20  teeth, 

.277X20  =  5.554  inches,  which  is  the  circumference 
of  the  pitch  circle 
and        5.554  +  K=l-768,   the   diameter  of  pitch   circle. 

Two    addendums=|-  inch,    which    added    to    D    makes 
OD  =  1.768 +  .125  =  1.893  inches. 

When  the  velocity  ratio  of  two  spiral  gears  is  other  than 
1  to  1,  it  cannot  be  determined  from  the  pitch  diameters, 
as  in  spur  gearing,  but  must  be  calculated  from  the  helical 
angle  of  the  teeth  of  each  gear.  If  the  helical  angle  is  the 
same  in  each  the  velocity  ratio  will  be  inversely  as  the  pitch 
diameters,  but  if  the  helical  angles  are  not  the  same,  the  num- 
ber of  teeth  per  inch  will  vary.  In  the  case  of  a  single  pitch 
worm  and  wheel,  the  worm  is  a  spiral  gear  of  one  tooth,  the 
velocity  ratio  being  1  to  the  number  of  teeth  in  the  wheel; 
increasing  or  decreasing  the  pitch  diameter  of  the  worm  will 
change  the  helical  angle  of  the  teeth  in  both  worm  and  wheel, 
but  will  not  affect  the  velocity  ratio  so  long  as  the  number 
of  teeth  remains  the  same  in  the  wheel. 


Note.  The  term  helix  is  used  to  denote  the  path  of  a  point 
moving  parallel  to  and  at  equal  distances  from  the  axis  of  a  cylin- 
der, while  the  spiral  is  a  point  moving  in  a  path  that  gradually 
increases  its  distance  from  the  axis.  As  the  helix  is  a  special  case 
of  the  spiral  it  is  customary  to  class  helical  work  as  spirals.  The 
angle  Y  which  the  spiral  makes  with  the  axis  of  the  gear,  is  called 
the  spiral  angle. 


SHOP  MATHEMATICS  45 

The  velocity  ratio  in  all  cases  depends  upon  the  number 
of  teeth  and  their  helical  angle,  from  which  is  derived  the 
formula  : 

v  :  V  =  DX  cos  A  :  d  X  cos  a 
V  =  velocity  of  large  gear. 
v  =  velocity  of  small  gear. 
D  =  pitch  diameter  of  large  gear. 
d  =  pitch  diameter  of  small  gear. 
A  and  a  =  the  angles  that  the  teeth  make  with  the  axes  of 

the  respective  gears. 

From  the  above  formula  the  velocity  ratio  of  two  spiral 
gears  is  proportional  to  the  pitch  diameters  only  when  the 
spiral  angles  of  the  gears  are  the  same,  or  45°  when  axes  are 
at  90°  to  each  other.  The  sum  of  the  spiral  angles  of  the  two 
gears  must  always  equal  the  angle  between  the  shafts,  and 
if  the  end  thrusts  on  the  bearings  of  the  gears  are  to  be  the 
same  the  angle  of  the  teeth  must  be  the  same  in  both.  When 
the  angle  of  the  spiral  to  the  axis  of  one  gear  is  greater  than 
the  other,  that  gear  should  be  the  driver. 

Example.  Two  spiral  gears  with  axes  at  90°,  velocity  ratio 
2  to  3,  are  to  be  cut  with  a  16  pitch  cutter.  The  pitch  dia. 
of  the  driving  gear  is  1^  in.,  and  the  teeth  are  at  an  angle  of 
60°  with  the  axis.  Calculate  the  other  dimensions  of  the 
gears. 

Solution.     The  normal  circular  pitch  of  driver 


Ib 

-    * 


Then 


46  SHOP  MATHEMATICS 

The  velocity  ratio  is  2  to  3, 
then       2  :  3=12  :  X.\  X  =  18,  the  number   of   teeth    in 

follower. 

To  find  the  normal  circular  pitch  of  follower 
Pt  =  .  19635, 

_  .19635  _ -19635^ 
cos  30°      .86603 


TT  7T 

The  outside  diameters  of  the  gears  are  found  by 

2 
adding  -=•  to  each  D,  the  resulting  blank  diameters  being 

1. 298 +%  inch  =  1.423 
and        1.5+^  =  1.625. 

The  formula  for  finding  the  number  of  the  cutter  to  use 
for  the  teeth  of  spiral  gears  is, 

No.  of  teeth  in  gear 
cube  of  the  cosine  of  the  spiral  angle 
In  the  above  example  it  is 

12          12 

:  =  — —  =  96  tooth  cutter 


cos3  60°     .125 

and  18          18 

-=  —  =  28  tooth  cutter. 


cos3  30P     .65 


PROBLEMS 


1.  Two  helical  gears  are  required  with  a  ratio  of  1  to  1, 
the  shafts  to  be  at  90°  to  each  other,  a  5  pi.  cutter  to  be 
used  and  the  OD  to  be  as  near  4  in.  as  can  be  obtained. 
Find  their  diameters. 

2.  Two  gears  with  shafts  parallel  to  each  other  are  to 
be  cut  with  a  10  pi.  cutter  and  to  have  10  teeth.    What  are 
the  sizes  for  the  helical  gear  with  a  velocity  ratio  of  1  to  1  ? 


SHOP  MATHEMATICS 


47 


3.  Find  the  blank  diameter  of  two  helical  gears  to  have 
12  teeth  when  the  velocity  ratio  is  1  to  1,  teeth  to  be  cut 
with  an  8  pi.  cutter  and  shafts  at  90°. 

4.  What  will  be  the  distance  between  the  centers  of  two 
helical  gears  with  velocity  ratio  of  1  to  1,  each  having  20 
teeth  cut  with  a  12  pi.  cutter? 

5.  Find  centers  of  two  helical  gears  with  20  teeth,  and 
20  pi.  cutter,  with  a  velocity  ratio  of  1  to  1. 

6.  Two  spiral  gears  with  axes  at  90°  to  each  other  and 
velocity  ratio  1  to  2  are  to  be  cut  with  a  14  pi.  cutter;  the 
outside  diameter  of  driving  gear  is  1.9  in.  and  angle  of  teeth 
is  63°  26'  to  axis  of  gear.    Find  the  other  dimensions  for  the 
gears. 

PULLEYS 

Cast  iron  pulleys  should  not  run  much  over  a  mile  per  min- 


ute at  the  surface  of  rim.  or 


5280 
60 


=  88  feet  per  second. 

The  following  dimensions 
may  be  used  for  cast  iron 
pulleys. 

When  b,  Fig.  36  =  width 
of  belt  for  required  power, 
B  =  width  of  rim  = 

t<&+4). 

c=  crown  of  rim,  with 
radius  varying  from  three  to 
five  times  the  width  of  run. 

Note.  When  the  crown  of  pulley  is  tapered  each  way  from  center 
of  rim  instead  of  a  curved  surface,  the  taper  for  6  in.  wide  and  under 
is  $  in.  per  ft.  6  in.  to  12  in.  wide  is  i  in.  per  ft.  12  in.  to  18  in. 
wide  is  f  in.  per  ft. 


48  SHOP  MATHEMATICS 

D  =  diameter  of  pulley. 
S  =  thickness  of    belt,    calculated    from 
the  H.  P.  required,  usually  ^  inch. 
f°r  single  and  f  inch  for  double  belts. 
£_.      a-,  t  =  thickness  of  rim  at  edge  =  .73  +  .005D. 

ty.  <J/.  T  =  thickness  Of  rim  at  center  =St+C. 

2B 
L  =  length  of  hub  =  —  . 

o 

Dl  =  diameter  of  hub  =  2  X  diameter  of  shaft. 
N  =  number  of  arms  =  4  to  6,  up  to  60  inches  D. 
h  =  width  of  arm  at  center  of  pulley  and 

approximately  =  —      —  +i  inch. 
ID 


h=  .633  •*/___  for  single  and 


h=  .798  -\-  for  double  belts. 
—  =  width  of  arm  at  rim  of  pulley. 

/£ 

e  =  .4h  for  elliptic  section  x  and  .5h  for  segmental 

section  F,  Fig.  37. 

The  H.  P.  that  a  pulley  will  transmit  can  be  calculated 
from  the  same  formula  as  the  one  for  belts  provided  that  it 
has  been  designed  with  the  right  proportions. 
The  following  formula  is  the  one  generally  used  : 
„  CXR.  P.M.Xb 


where    6  =  width  of  belt  in  inches. 

C=  circumference  of  pulley  in  feet. 

Note.     Diameters  of  pulleys  are  designed  to  be  not  less  than  18 
times  the  thickness  of  the  belt  that  is  to  run  over  them. 


SHOP  MATHEMATICS  49 

PROBLEMS 

1.  A  main  line  shafting  runs  170  R,  P.  M.;  a  driver  36 
in.  dia.  is  belted  to  12  in.  pulley  on  counter;   a  16  in.  driver 
on  same  counter  is  belted  to  4  in.  pulley  on  grinder.    What 
is  the  R.  P.  M.  of  grinder  spindle? 

2.  A  main  line  running  150  R.  P.  M.,  has  a  42  in.  driver 
belted  to  an  8  in.  follower  on  first  counter;   an  18  in.  driver 
on  first  counter  to  a  6  in.  follower  on  second  counter;   a  16 
in.  driver  on  second  counter  to  a  3  in.  pulley  on  machine 
spindle.    Find  the  R.  P.  M.  of  spindle. 

3.  Compute  the  maximum  thickness  of  belt  that  should 
be  used  on  the  machine  in  problem  2. 

4.  A  driving  pulley  30  in.  dia.  makes  150  R.  P.  M.    What 
is  the  size  of  the  follower  making  175  R.  P.  M.f 

5.  If  a  follower  is  6  in.  dia.  and  makes  1,000  R.  P.  M. 
what  is  the  size  of  driver  making  150  R.  P.  M.f 

6.  Compute  the  width  of  pulley  face  for  a  3  in.  belt. 

7.  What  is  the  width  of  pulley  for  a  7  in.  belt? 

8.  What  is  the  width  of  belt  for  a  pulley  13  in.  wide? 

9.  Find  c,  t,  and  T  when  S=$  in.,  Z)  =  6  in.  and  5  =  4  in. 

10.  Find  h  for  60  in.  D  and  14  in.  B  for  single  belt  where 
X  =  6. 

11.  A  pulley  is  60  in.  dia.  by  16  in.  face,  for  a  double  belt, 
4  in.  dia.  shaft  and  elliptic  cross  section  for  the  arms.    What 
are  the  dimensions  for  b,  c,  S,  N,  h,  e,  t,  T,  Dv  and  Lf 

MISCELLANEOUS  PROBLEMS 

1.  A  helical  gear  with  50  teeth  is  to  be  cut  with  a  20  pi. 
Brown  and  Sharpe  cutter.  What  are  the  blank  and  pitch 
diameters  and  lead  of  helix,  when  helical  angle  is  45°? 


50 


SHOP  MATHEMATICS 


2.  The  angle  of  a  helical  gear  is  45°  and  is  to  have  30 
teeth  cut  with  B.  &  S.  10  pi.  cutter.    What  is  the  D,  OD  and 
lead  of  helix  calculated  on  pi.  circle? 

3.  A  helical  gear  is  to  have  25  teeth,  and  45°  angle,  and 
to  be  cut  with  a  B.  &  S.  12  pi  cutter.     Find  D,  OD  and  lead 
of  helix  at  bottom  of  teeth. 

Note.  For  best  results  in  cutting  these  gears  it  has  been  found 
that  the  lead  of  helix  should  be  calculated  at  the  bottom  of  tooth, 
and  not  on  the  pitch  circle. 

4.  A  double  thread  worm,  Fig.  34,  with  16  in.  crank, 
meshes  with  a  40  tooth  wheel  16  in.  dia.    Fast  to  worm  wheel 
is  a  3  in.  dia.  axle.    When  a  force  of  75  Ibs.  is  applied  at  x, 
what  weight  can  be  raised? 

5.  The  same  as  problem  4  except  a  triple  thread  worm 
in  mesh  with  75  tooth  worm  wheel. 

6.  The  same  as  problem  4  except  a  single  thread  worm 
in  mesh  with  80  tooth  worm  wheel. 

7.  In  Fig.  38,  S  is  a  screw  of  f  in. 
lead,  B  a  gear  with  75  teeth  fast  to  S, 
A,  a  gear  of  15  teeth  fast  to  a  crank  C 
12  in.  long.  When  a  pull  of  75  Ibs. 
is  applied  at  x,  what  weight  can  be 
raised  at  W? 
riiil  8.  With  a  single  pitch  screw, 

__£ ^j  Fig.  38,  having  \  in.  lead  fast  to  gear 

B  with  80  teeth,  and  driven  by  gear 
A  with  20  teeth  that  is  fast  to  a  crank 
C   15  in.   long,   what  weight   can  be 
raised  at  W  with  a  pull  at  x  of  100  Ibs.? 

9.     The  same  as  problem  8,  except  with  a  double  pitch 
screw  and  50  Ibs.  pull  at  x. 


SHOP  MATHEMATICS 


51 


?  Nut 


E 


u f '--  - 

\^> 


10.  The  same  as  problem  8,  except  follower  gear  has  100 
teeth  and  driver  10  teeth 

and  crank  25  in.  long. 

11.  The  same   as  prob- 
lem 8,  except    W  is  9,975 
Ibs.  to  find  P. 

12.  In   Fig.  39,  S  is  a 
screw  of  }  in.  lead,  gear  E, 
fast  to  S,  has  100  teeth,  D 
has  40  teeth,  gears  B  and 
D  fast  to  same  stud,  and  B 
has    120   teeth,   A  has   30 
teeth  fast  to  crank  C,  27  in. 

long.     Find  W  when  the  force  applied  at  x  is  60  Ibs. 

13.  The  same  as  problem  12,  except  gears  D  and  B  have 
30  and  150  teeth  respectively. 

14.  What  power  applied  at  x,  problem  12,  will  raise  a 
weight  of  250,000  Ibs.? 

15.  In  Fig.  40,  C 
is  a  crank  12  in.  long 
fast  to  pulley  A  8  in. 
dia.  A  is  belted  to  B 
which  is  6  in.  dia.  D 
is  a  gear  of  10  teeth 
fast  to  B,  and  D  is 
in  mesh  with  gear  E 
having  50  teeth;  screw 
S  is  ^  in.  lead,  fast  to 
gear  E.  What  weight 
can  be  moved  up  in- 
cline L,  which  is  12  in.  long,  8  in.  high,  when  a  force  of  75 
Ibs.  is  applied  at  x? 


52  SHOP  MATHEMATICS 

16.  Find  the  force  that  will  be  required  at  x,  Fig.  40,  to 
move  a  weight  of  75,000  Ibs.  up  L,  when  S  is  f  in.  lead,  and 
C  is  16  in.,  pulleys,  gears  and  incline  same  as  in  problem  15. 

17.  If  the  positions  of  pulleys  A  and  B,  Fig.  40,  were 
exchanged  and  C  24  in.  long,  otherwise  same  as  in  problem 
16,  what  force  will  be  required  at  x? 

18.  If  a  single  thread  worm  was  used  in  place  of  gear  D 
and  belted  pulleys  A  and  B  of  Fig.  40,  and  a  12  in.  crank 
was  fast  to  worm  which  meshes  with  a  50  tooth  wheel  16  in. 
dia.  at  E,  L  and  S,  the  same  as  in  problem  15,  what  weight 
would  be  moved  up  L  with  force  of  75  Ibs.  at  end  of  crank 
on  worm? 

19.  If  in  Fig.  40,  L  is  8  ft.  and  H  is  3  ft.  and  S  is  i  in. 
lead  double  thread,  gears  E  and  D  15  in.  and  6  in.  diameters 
respectively,  and  A  and  B  15  in.  and  12  in.  diameters  re- 
spectively, what  force  applied  at  x  of  a  12  in.  crank  will 
move  60,000  Ibs.  up  the  incline  at  W? 

FRICTION 

Friction  is  the  resistance  offered  a  body  moving  on  a 
surface.  It  is  sliding  friction  when  one  surface  slides  on  the 
other,  and  rolling  friction  when  one  body  rolls  on  the  other 
so  that  new  surfaces  are  continually  coming  into  contact. 

A  wagon  moving  along  a  road  illustrates  rolling  friction 
between  the  wheels  and  roadway,  and  sliding  friction  be- 
tween wheels  and  axles. 

Sliding  friction  varies  greatly  according  to  the  materials 
used.  For  example,  a  sleigh  drawn  over  bare  ground  has 
more  friction  than  when  drawn  over  ice. 

/  is  the  symbol  used  to  designate  the  coefficient  of  friction. 
It  is  the  ratio  between  the  force  required  to  overcome  the 


SHOP  MATHEMATICS  53 

resistance  due  to  friction,  and  the  weight  or  pressure  of  the 
moving  body  on  a  horizontal  plane.     From  this  is  obtained 

/=|-,  and  P=/XW;       W=~. 

Any  pressure  at  right  angles  to  the  line  of  a  moving  body 
may  be  considered  as  part  of  its  weight. 

Example.  The  slide  valve  of  a  steam  chest  may  weigh 
only  15  Ibs.,  but  a  steam  pressure  of  100  Ibs.  per  sq.  in.  on 
the  valve  may  bring  a  weight  of  5,000  Ibs.  on  the  valve  seat 
to  resist  the  sliding  of  the  valve,  so  that  pressure  and  weight 
may  thus  be  equivalent  terms.  If,  in  this  example, 
f=.10,  then  fX 5,000=  .10 X 5,000  =  500  Ibs.,  which  will  be 
the  force  required  to  slide  the  valve  on  its  seat. 

Example.  If  a  block  of  granite  weighs  6,000  Ibs.  and 
/=  .1666,  what  force  will  be  required  to  slide  the  block  along 
a  platform? 

Solution.     6,000 X .  166  =  999  Ibs. 

Example.  If  the  weight  of  a  block  is  300  Ibs.  and  the 
force  necessary  to  slide  it  along  is  50  Ibs.,  find/. 

Solution.    f=  50 +  300  =.166. 

Axle  friction  is  sliding  friction,  but  the  bearings  are  made 
very  smooth  and  /  is  much  smaller  than  for  ordinary  sliding 
friction. 

The  value  for  /  on  well  lubricated  bearings  is  from  .01  to 
.05,  when  not  well  lubricated /is  usually  taken  at  .075. 

The  values  for  /  on  well  lubricated  flat  slides  have  been 
established  by  experiment  as  .08  to  .10,  when  not  so  well 
lubricated /is  given  as  .16  to  .20. 

The  friction  between  a  shaft  bearing  and  its  journal  is 
axle  friction. 


54  SHOP  MATHEMATICS 

Example.  If  a  turning  lathe  with  driving  spindle  3  in. 
dia.  making  50  R.  P.  M.  has  a  pressure  on  bearings  of  400 
Ibs.,  what  power  will  be  required  to  run  the  lathe  with  a  24 
in.  drive  pulley  when/=  .08? 

Solution.     The  resistance  due  to  friction  is 

400 X. 08  =  32   Ibs.,    and    the    force    necessary    to 

,  .  ,.  32X1% 

overcome  friction  =  — — — =  4  Ibs. 
l& 

The  travel  of  the  force  at  the  rim  of  a  24  in.  dia.  pulley  is 
50X2irr 


12 
and  the  power  required  is 

314-16X4    1,256.64 


=  50X6.2832  =  314.16  ft.  per  min. 
ft.-lbs.  =  .038  H.  P. 


33,000  33,000 
Then  the  power  necessary  to  overcome  friction  in  fly 
wheels  or  other  heavy  axle  bearings  is  found  by  taking  the 
product  of  the  frictional  resistance  and  the  distance  of 

i    r  ±u-    t  r\    u   r>      ForceX travel  . 

travel  of  this  force.    Or  H .  P.  = ^7^7; •    The  following 

00,000 

formula  can  be  taken  as  a  close  approximation  to  find  the 
H.  P.  absorbed  by  friction  in  heavy  shaft  bearings: 
„  p  =  WXfXNXvXd 

33,000X12 
The  above  formula  becomes 

H.P.  =  W  XfX  NxdX  .000008. 
W  =  load  on  bearing  in  pounds. 
d=dia.  of  shaft  in  inches. 
N=R.  P.  M. 


SHOP  MATHEMATICS  55 

PROBLEMS 

1.  A  fly  wheel  for  a  steam  engine  weighing  6,000  Ibs. 
makes  74  R.  P.  M.    Find  power  absorbed  by  friction  if  the 
fly  wheel  shaft  is  10  in.  dia.  and/=  .08. 

2.  A  fly  wheel  weighs  12  tons,  the  shaft  at  bearing  is  12 
in.  dia.,  makes  72  R.  P.  M.  and  f=.08.     Find  power  ab- 
sorbed by  friction. 

LAWS  OF  FRICTION 

1.  Friction  is  in  direct  proportion  to  the  pressure  with 
which  bodies  bear  against  each  other. 

2.  Friction  depends  upon  the  quality  of  the  surfaces  in 
contact. 

3.  Velocity  within  ordinary  limits  has  no  influence  on  the 
value  off. 

4.  The  area  of  surfaces  of  contact  does  not  affect  the  value 
of  f  within  ordinary  limits,  but  if  the  surfaces  are  unpropor- 
tionately  large  or  small  the  friction  will  be  increased,    f  for  roll- 
ing friction  such  as  that  between  the  car  wheels  and  rails, 
on  railways,  is  .002,  for  iron  tired  wheels  on  hard  roads  .02, 
on  soft  roads,  .06. 

ANGLE  OF  FRICTION  OR  REPOSE 

The  angle  which  a  plane  makes  with  the  horizontal  when 
a  body  just  begins  to  slide  on  that  plane  is  called  the  angle  of 
friction.  It  can  be  demonstrated  that  /  is  equal  to  the  tan- 
gent of  the  angle  of  friction. 

FRICTION  IN  PULLEY  BLOCKS 

Motion  between  two  bodies  in  contact  always  produces 
friction,  and  in  determining  the  efficiency  of  any  appliance 


56  SHOP  MATHEMATICS 

or  machine  the  friction  must  be  taken  into  account.  In  a 
simple  bar  or  lever  there  will  be  very  little  force  lost  in  fric- 
tion, but  in  the  pulley  there  is  the  friction  of  the  rope  bend- 
ing over  the  sheaves  as  well  as  the  friction  of  the  axles  in 
the  blocks. 

The  amount  of  force  required  to  overcome  friction  can  be 
readily  found  by  experiment.  The  following  formula  has 
been  found  to  be  right  for  the  effective  pull  obtained  by 
use  of  pulley  blocks.  When  W  is  the  weight  that  can  be 
raised  by  blocks  as  arranged  in  Fig.  18, 


according  to  the  number  of  sheaves  in  both  blocks;   and  for 
pulleys  as  arranged  in  Fig.  17, 


The  force  required  to  overcome  the  friction  of  a  body 
moving  up  an  inclined  plane  and  to  balance  the  weight,  by 
formula,  page  23,  will  be 


„  ^    .  ... 

F  =  —  f  --  h  friction. 
Li 

By  trigonometry  this  becomes 
F=  WXsin  A  +  friction; 

but  the  friction  is  equal  to  the  perpendicular  pressure  of  the 
body   on   the    inclined   plane 
multiplied   by   the  coefficient 
of  friction. 

Let  the  pressure  of  a  body 
on  an  inclined  plane  be  repre- 
sented by  line  1  —  3,  Fig.  41, 
perpendicular  to  AB,  while 
the  whole  weight  of  the  body 
will  be  represented  by  the  line 
1  —  2  perpendicular  to  the  base  AC  of  the  plane. 


SHOP  MATHEMATICS  57 

Draw  2  —  3  parallel  to  AB  and  perpendicular  to  1  —  3,  to 
complete  the  triangle.     Then  2  —  3  represents  the  force  re- 
quired to  balance  W  on  the  plane. 
2—3=WXsin  A 
l—3=WXcos  A 

since  triangle  123  is  similar  to  triangle  ABC  .'.  Friction  = 
WXcos  AXf,  &ndF=WXcos  Axf+WXsinA. 

Example.  A  weight  of  300  Ibs.  rests  on  a  plane  inclined 
at  30°  to  the  horizontal;  what  force  will  be  required  to  bal- 
ance the  weight  on  plane? 

Solution.     Friction  not  considered, 

F=WXsinA=300X.5  =  150  Ibs. 
Now  the  perpendicular  pressure  on  AB,  Fig.  41,  is 

300  X  (cos  A).  86603  =259.  8  Ibs., 
and  the  total  force  required  to  pull  W  up  the  incline  if  /=  .15, 

is        259.  8  X.  15  +150  =  188.  97  Ibs. 

If  W  is  moving  down  the  incline  then  the  force  due  to 
WXsin  A  will  help  to  move  it.  If  WXsin  A  is  more  than 
W  X  cos  A  Xf,  the  body  will  slide  of  itself.  Thus  in  the  above 
example  it  will  require  150  Ibs.  —  38.97  Ibs.  =  111.  03  Ibs.  to 
keep  the  weight  from  sliding. 

When  the  force  acts  parallel  to  the  base  of  the  incline,  as 
Plf  Fig.  41,  as  it  does  in  screw  threads  and  helical  cams, 
F=WXtanA  and  when  both  weight  and  friction  are 
considered  the  formula  is  : 


.  ,    »  .  A  \  * 

cos  A  —  (fXsin  A) 
Example.     When  W  =  300  Ibs.,  A  =30°,  f=  .15,  find  F. 

F-300X     - 
(  ~ 
238.5  Ibs. 


58  SHOP  MATHEMATICS 

The  mechanical  advantage  for  the  screw  has  been  given, 
as  P  :  W  =  Lead  :2-jrR,  but  the  screw  is  an  inclined  plane 
of  which  the  hypotenuse  is  the  middle  circumference  of 
the  thread,  and  the  lead  is  the  altitude  of  the  triangle,  with 
the  force  acting  parallel  to  the  base;  hence  when  friction  is 
considered  the  formula  is  : 

p_w      Lead  +  (fX^r)       r 

*  2irr—(fxLead)  A  R' 
where         F  =  total  force. 
W=  weight. 

R  =  radius  through  which  power  acts.  ' 
r=  middle  radius  of  screw  thread. 
2irr  =  middle  circumference  of  screw  thread. 
For  V  thread  screw,  the  frictional  resistance  will  be  in- 

creased as  -  T  =  the  secant  of  half  the  angle   of  thread, 
cos  A 

which  for  U.  S.  S.  thread  =  sec  of  30°  =  1.15.     The  formula 
for  V  threads  is 


._       r 

^  2-xr—  (LX/X1.16)       R} 
where  the   letters  have  the  same  meaning   as  in  formula 
above  for  square  threads. 


MISCELLANEOUS  PROBLEMS 

1.  What  force  will  be  required  to  raise  a  weight  of  1  ton 
with  a  double  sheave  pulley  as  in  Fig.  18? 

2.  What  force  will  be  required  to  raise  1,675  Ibs.  with  a 
double  sheave  pulley  as  arranged  in  Fig.  17? 

3.  How  much  extra  weight  would  a  180  Ibs.  man  require 
to  raise  2,000  Ibs.  with  a  3  sheave  pulley  as  arranged  in 
Fig.  17? 


SHOP  MATHEMATICS  59 

4.  What  is  the  value  of  /  when  the  pull  is  75  Ibs.  and 
the  weight  is  600  Ibe.? 

5.  What  is  the  value  of  /  when  the  pull  is  80  Ibs.  and 
the  weight  is  1,000  Ibs.? 

6.  What  is  the  pull  when  /is  .15  and  the  weight  is  800 
Ibs.? 

7.  What  force  will  be  required  to  haul  a  machine  weigh- 
ing 1,500  Ibs.  up  an  incline  with  the  horizontal  length  12  ft. 
and  height  5  ft.  when/ is  .2? 

8.  What  force  will  be  required  to  roll  a  cast  iron  cylinder 
weighing  1,150  Ibs.  up  an  incline  12  ft.  long  and  6  ft.  high 
when /is  .01? 

9.  An  engine  fly  wheel  and  shaft  10  in.  dia.  weighs  7 
tons.     What  is  the  power  required  to  move  wheel  when  / 
is  .08  and  shaft  makes  62  R.  P.  M.f 

10.  A  truck  loaded  with  a  box  of  castings  weighs  875  Ibs. 
If  the  two  bearings  of  truck  wheels  l\  in.  dia.  have/=.15, 
and  rims  of  wheels  have  /=  .02  on  floor  of  shop,  what  pull 
will  be  required  to  move  truck? 

11.  A  5  in.  pi.  dia.  worm  and  a  2  pi.  worm  wheel  of  36  teeth 
are  used  to  lift  an  elevator  weighing,  with  the  maximum  load, 
2  tons;    the  drum,  carrying  the  load  is  30  in.  dia.  and  is 
fastened  to  worm  wheel  shaft  which  makes   10  R.  P.  M. 
The  shaft  bearings  are  3  in.  dia.  with/=  .05;  the /for  sliding 
of  teeth  of  worm  and  wheel  is  .15  and  for  step  bearing  3  in. 
dia.  /= .  10.     WThat  force  will  be  required  at  the  rim  of  an 
18  in.  dia.  pulley  on  worm  shaft  to  start  the  elevator? 

12.  If  an  elevator  weighing  475  Ibs.  is  to  be  raised  a  dis- 
tance of  9  ft.  with  4  sheave  pulley  blocks,  what  pull  will  be 
required,  and  what  is  the  movement  of  the  rope,  with  pulley 
arranged  as  in  Fig.  17? 


60  SHOP  MATHEMATICS 

13.  How  much  force  will  be  required  to  roll  a  barrel  of 
sugar  weighing  300  Ibs.  up  a  plank  8  ft.  long  inclined  at.  15° 
when  /=.  005? 

14.  What  force  will  be  required  at  the  end  of  a  26  in. 
lever  to  raise  20,000  Ibs.  with  a  square  thread  screw  3  in.  dia. 
and  |  in.  lead  when /=  .15? 

15.  What  weight  can  be  lifted  with  a  square  thread 
screw  of  £  in.  lead  2^  in.  dia.  with/=.15,  when  a  force  of 
25  Ibs.  is  put  on  end  of  a  20  in.  lever? 

16.  25  tons  weight  is  to  be  lifted  with  50  jack  screws  of 
f  in.  lead,  and  3  in.  dia.  of  screw,  with/=.10.    What  force 
will  be  required,  distributed  evenly  at  ends  of  12  in.  levers? 

17.  A  hand  screw  press  is  used  to  punch  holes  in  a  sheet 
of  iron,  the  pressure  required  being  1,250  Ibs.     What  force 
will  be  required  at  the  end  of  a  16  in.  lever  on  a  triple  square 
thread  screw  2£  in.  dia.  f  in.  lead  and/=.25,  to  punch  the 
holes? 

18.  A  £  in.  lead  square  thread  screw  2  in.  dia.  with  /=  .10 
is  used  to  pull  a  broach  through  the  £  in.  square  hole  of  an 
end  wrench.    A  12  in.  pulley  with  200  Ibs.  pull  at  the  rim 
was  required  to  revolve  the  screw.    What  was  the  resistance 
of  the  pull? 

19.  A  hand  press  with  screw  2^  in.  dia.  and  ^  in.  lead  V 
thread  is  used  to  force  small  pins  into  small  discs.    A  pull  of 
40  Ibs.  is  used  on  the  rim  of  a  16  in.  hand  wheel.    If /=.15, 
what  is  the  resistance  against  the  pull? 

20.  If  100  ft.  of  shafting  2  in.  dia.  equipped  with  the 
usual  number  of  drive  pulleys  and  belting,  at  100  R.  P.  M. 
requires  1  H.  P.  to  keep  it  moving,  what  H.  P.  will  be  re- 
quired to  move  the  same  shaft  at  170  R.  P.  M.,  when  the 
power  required  increases  directly  as  the  speed? 


SHOP  MATHEMATICS  61 

21.  If  the  power  required  to  revolve  a  line  of  shafting 
increases  as  the  cube  of  its  dia.,  what  power  will  be  required 
to  turn  a  line  of  shafting  3  in.  dia.,  150  ft.  long  when  other 
requirements  are  the  same  as  in  problem  20? 

22.  What  power  is  absorbed  in  friction  in  a  lathe  spindle 
2^  in.  dia.  with  830  Ibs.  weight  on  bearings  due  to  belt 
pressure,  etc.,  on  a  10  in.  dia.  pulley  at  100  R.  P.  M.,  when 
/is  taken  at  .01? 

23.  What  is  the  power  absorbed  in  running  a  large  lathe 
with  a  10  in.  dia.  pulley  making  20  F.  P.  M.,  if  the  weight 
on  bearings  is  equal  to  3,600  Ibs.,  the  dia.  of  spindle  is  5  in. 
and /equals  .01? 

BELTING 

A  common  method  for  the  transmission  of  rotary  motion 
is  with  leather  belting  over  pulleys.  The  adhesion  of  a 
leather  belt  to  the  surface  of  a  pulley  rim  is  given  in  terms 
of  the  amount  of  weight  it  is  capable  of  lifting;  the  speed 
in  feet  per  minute  times  the  weight  lifted  equals  the  foot 
pounds  of  work  it  is  capable  of  doing. 

Cooper,  in  his  treatise  on  belting,  considers  a  safe  velocity 
for  leather  belts  to  be  fifty  square  feet  per  minute ;  for  a  belt 
one  inch  wide  this  would  be  equivalent  to  a  linear  velocity 
of  600  F.  P.  M. 

The  breaking  strain  for  leather  belting  is  given  as  3,200 
pounds  per  square  inch  of  cross  section;  using  a  factor  of 

3  200 
safety  of  11,  -j—  =  290  Ibs.  =the  safe  strain.   Nowfor  belting 

^  inch  thick  and  1  inch  wide,  290  X^—55  Ibs.  strain  per  inch. 
Double  belts,  that  is,  two  single  belts  cemented  together 


62  SHOP  MATHEMATICS 

may  have  about  one-half  more  added  for  the  allowable  pull 
on  the  belt,  or  82  £  pounds  per  inch  of  width. 

Modern  shop  practice  allows  less  strain  on  the  belt  than 
that  given  above,  as  low  as  30  pounds  being  given  by  some 
belt  makers  and  users.  This  would  correspond  to  a  linear 
velocity  of  over  1,100  F.  P.  M.  per  H.  P.  for  each  inch  of 
width,  but  about  800  F.  P.  M.  per  H.  P.  is  a  fair  average  for 
single  belts,  and  550  F.  P.  M.  per  H.  P.  for  double  belts. 

In  selecting  the  size  and  speed  for  belts  general  shop 
conditions  must  be  taken  into  account. 

Belts  should  not  run  faster  than  6,000  F.  P.  M.  on  account 
of  reducing  the  tension  of  the  belt  on  the  pulley  by  centrif- 
ugal force;  between  3,000  and  4,000  F.  P.  M.  is  considered 
the  better  practice. 

The  speed  of  a  belt  is  found  by  multiplying  the  R.  P.  M. 
of  driver  by  trD  or  by  formula:  F.  P.  M.  =  R.  P.  M.X*D. 
D  is  diameter  of  driver  in  feet. 

In  the  calculations  for  speeds  of  pulleys  for  belt  transmis- 
sion, 2%  may  be  allowed  for  slip,  or  creep,  of  belt  on  the  pul- 
leys. 

The  following  rule  is  approximately  correct  for  lengths  of 
belts  over  driver  and  follower  pulleys. 

RULE  1  .    Add  twice  the  distance  between  centers  of  shafts 
to  half  the  sum  of  the  diameters  of  the  two  pulleys  multi- 
plied by  TT, 
or  by  formula: 


I  —  whole  length  of  belt. 
L  =  distance  between  centers  of  pulleys. 
D  =  diameter  of  larger  pulley. 
d=  diameter  of  smaller  pulley. 


SHOP  MATHEMATICS 


63 


,» 

v 


----[--HA] 
«*v 


The  exact  lengths  for  belts  for  two  mating  pulleys  are  found 
as  follows: 

RULE  2.  Find  arc  of 
contact  of  belt  on  driver 
and  follower  and  to  their 
sum  add  twice  the  dis- 
tance  between  points  of 
contact  on  driver  and 
follower. 

This  distance  may  be 
found  by  the  graphical 
method,  or  as  follows: 
cos  A     R  —  r 


. 

Fig.    42, 


on 


for    arc    of    contact 
small  pulley. 

L=  distance  between 

centers  of  shafts  in  inches. 
R  =  radius  of  large  pulley. 
r  —  radius  of  small  pulley. 
A  =  arc  of  contact  on  small  pulley. 
A±  =  arc  of  contact  on  large  pulley  =  360° — A . 
The  arc  of  contact  on  cross  belts,  Fig.  43,  is  the  same 
for  both  pulleys  and  formula  is : 

/..         A\      R+r 
cos  I  180 — 


The  length  of  arc  of  contact  of  belt  on  pulley  is  found  by 
the  formula 

T=A°C 
360' 
where    L  =  length  of  arc  of  contact. 


64  SHOP  MATHEMATICS 

A  =  angle  of  arc. 
C  =  the  circumference  of  pulley. 

The  following  approximate  formula  for  width,  of  belt  for 
a  given  H.  P.  may  be  taken  as  close  enough  for  practical  use 
for  single  belts. 

H.  P.  X  50,  000  .  .  H.  P.  X  50,  000 

b=  —  ;  —  ==  —  -^=  —  ,  by  transposing  a  =  —  =  —  ===  —  -^  — 
dxWxN  bXWxN 

•,uu    *uu-     •     u  IT/     H.P.X  50,000 

b  =  width  of  belt  m  inches,  W  =  —  -^  —  7-  —  ^  — 

a  /\  o  /\  i  v 

,     ,.  f      H      •    •     u         AT-     H.P.X50,000 

a  =  diameter  of  pulley  in  inches,    N  =  —  ,     ,       '  — 

&  X  o  X  i' 

W  =  weight  per  square  foot  in  ounces, 

^ 


50,000 
N=R.  P.  M. 

1  square  foot  of  leather  belt  ^  inch  thick,  weighs  16  ounces. 
H.  P.  X  67,  500. 


For  double  belts,  6  = 


dxWxN 


PRESSURE  ON  BEARINGS 

Let  P—  pressure  on  bearings  from  pull  of  belt. 
V= travel  of  belt  in  F.  P.  M. 

.      ,  .     D    3X33,OOOXH.  P. 
Then  approximately  P= • — ~ 

PROBLEMS 

1.  A  single  belt   ^    in.   thick,  12  in.  wide  runs  5,000 
F.  P.  M.    If  there  is  a  strain  of  55  Ibs.  per  in.  in  width,  what 
H.  P.  can  be  transmitted  by  the  belt? 

2.  A  double  belt  20  in.  wide  runs  over  a  4  ft.  dia.  pulley 
at  180  R.  P.  M.    Find  H .  P.  that  the  belt  will  transmit. 


SHOP  MATHEMATICS  65 

3.  A  single  belt  24  in.  wide  runs  on  14  ft.  dia.  pulley  at 
75  R.  P.  M.    Find  the  H.  P.  it  is  capable  of  transmitting. 

4.  Two  pulleys  60  in.  and  24  in.  diameters  respectively 
are  on  separate  shafts  12  ft.  between  centers.     What  is  the 
approximate  length  of  belt  required? 

5.  What  is  the  exact  length  of  a  cross  belt  running  over 
two  pulleys  60  in.  and  24  in.  diameters,  when  the  shafts  are 
16  ft.  between  centers? 

6.  How  many  H.  P.  can  be  transmitted  by  a  3  in.  belt 
at  55  Ibs.  strain  per  in.  of  width  at  3,500  F.  P.  M.? 

7.  How  many  H.  P.  can  be  transmitted  by  a  4  in.  belt, 
35  Ibs.  strain,  at  4,800  F.  P.  M.? 

8.  How   many   H.  P.  can  be  transmitted  by  a  24  in. 
double  belt,  over  an  8  ft.  driver  at  115  R.  P.  M.? 

9.  How  many  H.  P.  can  be  transmitted  by  a  26  in.  single 
belt  at  33  Ibs.  strain  on  a  14  ft.  driver,  at  80  R.  P.  MJ 

10.  If  the  belt  in  problem  9  is  running  over  a  6  ft.  fol- 
lower pulley,  16  ft.  between  centers,  what  is  the  weight  for 
a  belt  ^  in.  thick? 

11.  Find  the  length  of  a  cross  belt  16  ft.  between  the 
centers  of  pulleys  that  are  14  ft.  and  6  ft.  diameters  respec- 
tively? 

12.  Find  the  exact  length  for  an  open  belt   12   ft.    be- 
tween centers  of  pulleys  that  are  respectively  84  in.  and 
48  in.  diameters. 

13.  What   is   the   length  of  a  cross  belt  16  ft.  between 
centers  of  pulleys  that  are    respectively    7    ft.    and   3    ft. 
diameters? 

14.  Find  exact  length  for  an  open  belt   18  ft.   between 
the  centers   of   pulleys  that    are   respectively   30   in.    and 
70  in.  diameters. 


66  SHOP  MATHEMATICS 

15.  What  is  the  width  of  a  double  belt  to  transmit  112 
H.  P.  on  a  4  ft.  diameter  pulley  at  180  R.  P.  M.f 

16.  What  H .  P.  will  a  double  belt  30  in.  wide  weighing  36 
oz.  per  sq.  ft.,  transmit  on  a  60  in.  dia.  pulley  at  200  R.  P.  M.? 

17.  A  16  ft.  dia.  pulley  runs  at  75  R.  P.  M.;  over  this  is 
running  a  24  in.  double  belt  that  weighs  32  oz.  per  sq.  ft. 
What  H.  P.  will  the  belt  transmit? 

18.  What  is  the  pressure  on  bearings  of  a  belt  that  is 
running  900  F.  P.  M.  and  transmits  60  H,  PJ 

19.  What  will  be  the  pressure  on  the  caps  of  lathe  spindle 
bearings,  when  a  4  in.  belt  on  10  in.  dia.  pulley  making  100 
R.  P.  M.  does  2  H.  P.  of  useful  work? 

20.  What  is  the  pressure  on  the  bearings  for  a  4  in.  grinder 
belt  at  3,000  F.  P.  M.  transmitting  £  H.  P.? 

21.  What  is  the  pressure  on  the  bearings  of  a  drop  ham- 
mer pulley  with  5  in.  belt  running  at  a  velocity  of  2,000 
F.  P.  M.  when  5  H.  P.  is  to  be  transmitted? 

22.  What  dia.  of  pulley  would  be  required  to  increase 
the  velocity  of  a  belt  from  900  F.  P.  M.  over  a  pulley  20 
in.  dia.  to  3,000  F.  P.  M.  with  the  same  R.  P.  M.f 

23.  Find  the  H.  P.  of  a  24  in.  double  belt,  36  oz.  weight 
per  sq.  ft.  with  36  in.  dia.  drive  pulley  making  155  R.  P.  M. 

24.  Find  H.  P.  of  40  in.  double  belt  with  4  reinforcing 
strips  each  4  in.  wide.    The  strips  weigh  16  oz.  per  sq.  ft.,  the 
belt  32  oz.  per  sq.  ft.  and  runs  on  a  20  ft.  fly  wheel  pulley 
at  65  R.  P.  M. 

25.  What  is  the  H.  P.  of  a  16  in.  single  belt  over  an  8  ft. 
drive  pulley  at  110  R.  P.  M.f 

26.  What  is  the  H.  P.  of  a  24  in.  double  belt  on  48  in. 
driver  at  275  R.  P.  M.  when  the  weight  is  24  oz.  per  sq.  ft.? 


SHOP  MATHEMATICS  67 

27.  What  width  of  double  belt  £  in.  thick  will  be  required 
for  a  28  H.  P.  drive  at  3,500  F.  P.  M.? 

28.  What  width  of  double  belt  at  27  oz.  per  sq.  ft.  on  a 
24  in.  dia.  driver  at  150  R.  P.  M.  will  be  required  for  trans- 
mitting 62  i  H.  P.  ? 

29.  Find  b  when  w=  16,  N=  125,  d=20  in.,  H.  P.  =  50. 

30.  Find  u>  when  6  =  20  in.,  d  =  36  in.,  N=2QO,  H.  P.  =  75. 

31.  What  weight  of  belt  per  sq.  ft.  will  be  required  to 
transmit  250  H.  P.  over  a  72  in.  dia.  pulley  at  250  R.  P.  M., 
when  belt  is  36  in.  wide? 

32.  What  R.  P.  M.  will  a  34  in.  dia.  drive  pulley  make 
to  transmit  100  H.  P.  with  a  24  in.  belt  weighing  32  oz.  per 
sq.  ft.? 

33.  What  R.  P.  M.  will  a  96  in.  dia.  drive  pulley  be  re- 
quired to  make  in  order  to  transmit  175  H.  P.  with  a  belt 
36  in.  wide  and  weighing  36  oz.  per  sq.  ft.? 

34.  Find  the  dia.  of  a  drive  pulley  making  125  R.  P.  M. 
for  a  28  in.  belt  weighing  32  oz.  per  sq.  ft.  to  transmit  185 
H.  P. 

35.  What  is  the  dia.  of  pulley  required  at  250  R.  P.  M. 
for  a  belt  30  in.  wide  weighing  32  oz.  per  sq.  ft.  to  transmit 
185  H.  P.? 

36.  What  weight  per  sq.  ft.  will  be  required  for  a  belt  12 
in.  wide  running  3,600  F.  P.  M.  on  a  24  in.  dia.  pulley,  to 
transmit  75  H.  P.? 

37.  What  weight  per  sq.  ft.  will  be  required  for  a  24  in. 
belt  on  a  46  in.  dia.  pulley  making  225  R.  P.  M.  to  transmit 
125  H.  PJ 

38.  Find  w  when  R.  P.  M.  =  750,  d=15  in.,  6  =  7  in., 
H.  P.  =  25. 


68  SHOP  MATHEMATICS 

ROPES 

Rope  drives  are  used  for  long  distance  transmissions  and 
for  drives  leading  off  to  several  shafts  at  different  angles 
to  each  other  and  not  parallel  to  the  driving  shaft.  The 
effectiveness  of  a  rope  drive  depends  upon  the  friction  in 
the  grooves  of  the  pulleys. 

The  velocity  of  rope  transmission  is  from  1,500  to  5,000 
F.  P.  M.  and  should  not  exceed  5,000  F.  P.  M.  for  then  the 
loss  due  to  centrifugal  force  will  offset  the  gain  in  speed. 
The  diameter  of  sheave  should  not  be  less  than  40  times  the 
diameter  of  rope  in  inches. 

The  breaking  strain  for  hemp  rope  is  6,000  pounds  per 
square  inch  of  cross  section  area;  for  manila  rope  3,000 
pounds  per  square  inch. 

To  find  the  breaking  strength,  multiply  the  cross  section 
area  of  the  rope  by  6,000  for  hemp  and  by  3,000  for  manila 
rope. 

The  horse  power  which  a  rope  is  capable  of  transmitting 
may  be  found  by  the  formula: 


..  .. 

oUUO 

d=  diameter  of  rope  in  inches. 
V=F.  P.  M. 

The  weight  of  1  inch  diameter  rope  is  T3^  of  a  pound  per 
foot  in  length,  the  weights  of  other  sizes  can  be  found  by 
formula: 

W=d2X.3. 

PROBLEMS 

1.  Find  the  weight  of  400  ft.  of  1J  in.  dia.  rope. 

2.  Find  the  weight  per  ft.  of  1  £  in.  dia.  rope. 

3.  What  weight  will  be  required  to  break  a  1|  in.  dia. 
manila  rope? 


SHOP  MATHEMATICS  69 

4.  What  weight  would  be  likely  to  break  a  f  in.  dia. 
hemp  rope? 

5.  What  weight  would  be  apt  to  break  a  If  in.  dia. 
manila  rope? 

6.  What  H.  P.  will  a  \\  in.  dia.  manila  rope  transmit, 
running  over  two  8  ft.  sheaves  at  150  R.  P.  MJ 

7.  If  the  pulleys  in  problem  6  are  100  ft.  apart  on  centers, 
what  is  the  weight  of  the  rope? 

8.  What  strain  will  be  put  on  a  If  in.  manila  rope  run- 
ning at  4,500  F.  P.  M.,  to  transmit  150  H.  P.? 

9.  What  H.  P.  will  be  transmitted  with  a  If  in.  manila 
rope  at  5,500  F.  P.  MJ 

10.  What  is  the  minimum  size  of  sheaves  and  R.  P.  M. 
of  pulley  for  a  If  in.  dia.  manila  rope  at  3,500  F.  P.  MJ 
Also,  what  H.  P.  can  be  transmitted  by  this  rope? 

11.  What  weight  would  be  required  to  break  a  f  in.  dia. 
manila  rope? 


WIRE  CABLE  TRANSMISSION 

With  a  wire  cable  drive  there  should  be  no  binding  or  wedg- 
ing of  the  cable  in  the  groove  of  the  sheave.  The  cable 
should  run  on  wood  or  leather  packing  in  the  bottom  of  the 
groove,  and  the  sheave  for  iron  wire  should  not  be  less  than 
150  times  the  diameter  of  the  cable.  The  cable  should  not 
run  over  6,000  F.  P.  M.  on  account  of  centrifugal  force, 
and  3,000  to  5,000  F.  P.  M.  is  preferable. 

The  distance  between  centers  of  pulleys  should  not  be 
less  than  60  feet  nor  over  400  feet.  Carrying  pulleys  are  used 
when  the  spans  are  over  400  feet. 


70  SHOP  MATHEMATICS 

The  H.  P.  transmitted  by  an  iron  wire  cable  is  found  by 
the  formula: 

=  d*XVX875 

5,000 

or    H.  P.  =  d2xVX.055, 

where  the  letters  have  the  same  significance  as  in  the  for- 
mulas for  rope  drives.  A  steel  cable  will  transmit  double  the 
above  amount. 

The  breaking  strain  for  iron  cables  is  40,000  pounds  per 
square  inch  of  cross  section  area,  and  for  steel  cables  80,000 
pounds. 

A  factor  of  safety  of  10  is  generally  allowed  for  rope  and 
cable  transmission. 

A  factor  of  safety  is  the  number  which  expresses  the  ratio 
of  the  ultimate  strength  of  a  body  to  the  working  load. 

The  weights  of  wire  cables  per  foot  in  length  are  as  follows : 

f   in.  dia.  =  .21  Ibs.  f  in.  dia.=   .57  Ibs. 

A  in.  dia.  =  .23  Ibs.  f  in.  dia.  =    .92  Ibs. 

\  in.  dia.  =  .31  Ibs.  |  in.  dia.  =  1.20  Ibs. 

1  in.  dia.  =  1.50  Ibs. 

Cable  Transmission  Problems 

1.  What  is  the  weight  of  a  f  in.  dia.  cable  at  1.2  Ibs.  per 
ft.  837  \  ft.  long,  and  what  H.  P.  can  be  transmitted  at  100 
R.  P.  M.  over  12  ft.  dia.  sheaves? 

2.  What  is  the  difference  in  H.  P.  of  problem  1  when 
figured  by  the  formula  and  by  ft.-lbs.  of  work? 

3.  What  is  the  weight  of  a  f  in.  iron  cable  700  ft.  long, 
and  what  H.  P.  can  be  transmitted  over  10  ft.  dia.  sheaves 
at  80  R.  P.M.  ? 


SHOP  MATHEMATICS 


71 


4.  What  H .  P.  can  be  transmitted  by  a  1  in.  cable  at  95 
R.  P.  M.  over  a  12 £  ft.  dia.  sheave? 

5.  The  length  of  a  1£  in.  iron  cable  is  800  ft.  and  it  weighs 
2  Ibs.  per  ft.     Find  the  F.  P.  M.  required  to  transmit  200 
H.  P. 

6.  Find  the  dia.  of  sheave  at  125  R.  P.  M.  and  length  of 
a  |  in.  steel  cable  to  transmit  90  H.  P.  and  be  inside  the  safe 
limit. 

CHAIN  TRANSMISSION 

One  form  of  chain  used  for  hoisting  machinery  is  made 
of  round  wire  links  as  short  as  possible  for  strength.  Let 
d  =  diameter  of  wire,  then  the  length  of  link  inside  =  2. 5  d; 
width  of  link  inside  =  1.5  d. 

The  following  table  gives  sizes  of  wire  for  hoists : 


Capacity  in  tons, 

i 

i 

i 

H 

2 

3 

4 

5 

6 

8 

10 

Diam.  of  wire  in  inches, 

A 

i 

A 

A 

f 

& 

i 

A 

I 

H 

« 

Another  form  of  chain  used  for  giving  a  positive  motion 
between  two  shafts  by  means  of  sprocket  wheels,  is  the  flat 
link  and  block  chain;  here  the  strain  on  the  chain  is  limited 
by  the  shearing  stress  on  the  pivots. 

The  sprockets  should  not  have  less  than  7 
teeth;  the  larger  the  sprocket  the  less  will 
-.  J   be  the  strain  on  chain  and  consequently  less 
/    wear  on  the  pivots  of  the  rivets. 

The  formula  for  sizes  of  sprockets  is  as 
follows : 

=  180° 
:~N~' 


72  SHOP  MATHEMATICS 


__         sm  x 
tan    Y  = 


B  +  cos  x 
A 


sin 

N-=  number  of  teeth  in  sprocket, 
A  =  distance  between  centers  of  rivets  in  link, 
B=  distance  between  centers  of  rivets  in  block, 
6  =  diameter  of  end  of  block, 

for  1  inch  pitch  of  chain  6  is  usually  .325  inch  when  A  = 
.6  inch  and  B=  A  inch. 

The  bottom  diameter  of  sprocket  wheel  is  the  important 
dimension,  therefore  size  b  must  be  taken  accurately. 
Then    OD  =  PD+b 
and      Bottom  diameter  =PD  —  6. 

PROBLEMS 

1.  What  is  the  dia.  of  an  8  tooth  sprocket  wheel  with  a 
1  in.  pi.  chain? 

2.  What  is  the  OD  and  bottom  dia.  of  a  20  tooth  sprocket 
with  1  in.  pi.  chain? 

3.  What  are  the  distances  A,  B,  and  dia.  ofb,  for  a  1^  in. 
pi.  chain,  when  proportionally  the  same  as  for  1  in.  pitch? 

4.  What  are  the  OD  and  PD  for  a  20  tooth  sprocket, 
with  1  £  in.  pi.  chain? 

5.  Find  the  diameters  for  a  24  tooth  sprocket  for  a  1£  in- 
pi.  chain. 

6.  Find  A,  B,  and  b,  for   If  in.  pi.  when  proportionally 
the  same  as  for  a  1  in.  pi.  chain. 

7.  Find  the  diameters  for  a  28  tooth  sprocket  with  a  If 
in.  pi.  chain. 


SHOP  MATHEMATICS  73 

8.  What  are  the  PD  and  OD  for  a  sprocket  of  28  teeth 
and  1  in.  pi.  chain? 

9.  Find  the  bottom  dia.  for  a  sprocket  of  30  teeth  and 
1  in.  pi.  chain. 

SHAFTING 

The  H.  P.  which  a  line  shaft  can  impart  to  connected 
machinery  is  limited  by  the  strength  of  the  material  of  which 
it  is  made.  The  principal  strain  is  in  the  twisting  of  the 
round  bar  when  the  pulleys  are  made  to  revolve  carrying 
driving  belts  to  the  various  machines. 

The  twisting  of  the  shaft  is  called  its  torsional  strain,  and 
the  formula  which  determines  the  amount  of  torsion  which 
a  shaft  will  safely  stand  is 
,      »/    rw 
= 


where  d  =  the  diameter  of  shaft  in  inches, 
w  =  the  pull  of  the  belt  in  pounds, 
r  =  the  radius  of  the  pulley  in  feet, 
C  =  the  constant  for  breaking  moment  which  is  found 
by  experiment  for  cold  rolled  steel  to  be  660 
pounds. 

10  is  the  factor  of  safety. 
Then  the  above  formula  may  be  written 
dss*IHX80 


N 
where  H  =  H.P. 

N  =  R.  P.  M. 


74  SHOP  MATHEMATICS 

PROBLEMS 

1.  Find  values  for  H  and  N  by  transposing  the  formula 
for  d,  given  above. 

2.  A  4  in.  dia.  shaft  runs  150  R.  P.  M.     What  is  the 
safe  load  in  H.  P.  to  put  on  the  shaft? 

3.  Find  the  H .  P.  that  can  be  transmitted  by  a  shaft  1\ 
in.  dia.  at  200  R.  P.  M. 

4.  What  H.  P.  can  be  transmitted  by  a  5  in.  dia.  shaft 
running  at  275  R.  P.  MJ 

5.  What  dia.  shaft  will  be  required  to  transmit  500  H.  P. 
running  at  150  R.  P.  MJ 

6.  What  dia.  of  shaft  will  be  required  to  transmit  250 
H.  P.  running  at  74  R.  P.  M.f 

7.  How  fast  should  a  shaft  revolve  that  is  to  transmit 
1,000  H.  P.  and  is  12  in.  diameter? 

8.  What  is  the  dia.  of  a  shaft  that  is  to  transmit  150 
H.P.&H50R.P.MJ 

9.  Find  the  R.  P.  M.  of  a  5  in.  shaft  to  transmit  300  H.  P. 

10.  Find  the  dia.  of  a  shaft  to  transmit  12}  H.  P.  at  500 
R.  P.  M. 

11.  Find  the  dia.  of  a  shaft  to  transmit  400  H.  P.  at  200 
R.  R.  M. 

12.  What  dia.  of  shaft  will  be  required  to  transmit  1,000 
H.  P.  at  36  R.  P.  M.f 

13.  What  dia.  of  shaft  will  be  required  to  transmit  200 
H.P.&tlSOR.P.M.f 

14.  Find  the  H.  P.  transmitted  by  an  8  in.  dia.  shaft  at 
115  R.  P.  M. 


SHOP  MATHEMATICS  75 

15.  Find  the  H.  P.  transmitted  by  a  6  in.  dia.  shaft  at 
150  R.  P.  M. 

16.  What  H.  P.  can  be  transmitted  by  an  18  in.  dia. 
shaft  at  50  R.  P.  M.f 

17.  How  many  R.  P.  M.  should  an  8  in.  dia.  shaft  make 
to  transmit  200H.PJ 

18.  How  many  R.  P.  M.  should  a  3  in.  dia.  shaft  make 
to  transmit  200  H.  P.? 

19.  How  many  R.  P.  M.  should  a  l}f   in.  dia.  shaft 
make  to  transmit  28  H.  P.? 

A  jack  shaft  is  a  short  shaft  between  the  engine  and  main 
line  of  shafting,  and  is  used  to  regulate  the  speed  of  the  main 
line,  and  prevent  the  heavy  strains  on  the  main  line 
shafting  due  to  the  engine  drive  belt.  The  bearings  must 
be  as  near  the  pulleys  as  possible. 

The  strain  on  shaft  being  greater,  the  constant  is  increased 

from  80  to  120.     The  formula  is  d=\HX*®°    the  same 

\      N 

letters  having  the  same  meaning  as  in  formula  for  shafting 
given  on  page  73. 

Idler  shafts  are  used  to  change  the  direction  in  which  the 
belts  run  and  have  only  a  bending  or  shearing  strain. 

The  distance  between  the  centers  of  bearings  for  line  shaft- 
ing varies  in  proportion  to  the  size  of  shaft. 


Dia.  of  shaft 

li  to  If  in. 

2  to  1\  in. 

2i  to  4  in. 

Center  of  boxes 

7  ft. 

8  ft. 

10  ft. 

76  SHOP  MATHEMATICS 

PROBLEMS 

1.  What   dia.   of  jack  shaft  will  be   required   at   225 
R.  P.  M.  to  transmit  225  H.  P.f 

2.  What  dia.  of  jack  shaft  will  be  required  at  500  R.  P. 
M.  to  transmit  250  H.  P.? 

3.  How  many  R.  P.  M.  should  a  6  in.  dia.  jack  shaft 
make  to  transmit  500  H.  P.f 

4.  How  many  R.  P.  M.  should  a  2  in.  dia.  jack  shaft 
make  to  transmit  87  £  H.  P.f 

5.  What  H.  P.  will  a  6  in.  jack  shaft  transmit  at  250 
R.  P.  M.f 

6.  What  H.  P.  will  a  2  in.  dia.  jack  shaft  transmit  at 
250  R.  P.  M.f 

7.  Find  H.  P.  transmitted  by  a  2£  in.  dia.  jack  shaft 
at  170  R.  P.  M. 

8.  Find  dia.  of  jack  shaft  designed  for  a  250  H.  P. 
engine,  when  the  shaft  is  to  make  84  R.  P.  M. 

9.  What  should  be  the  distance  between  the  centers  of 
hangers  for  a  line  shaft  that  will  transmit  100  H.  P.  at  160 
R.  P.  MJ 

10.  Find  the  distance  between  the  centers  of  hangers  on 
a  line  shaft  to  transmit  50  H.  P.  at  150  R.  P.  M. 

JOURNAL  BEARINGS 

The  lengths  of  journal  bearings  vary  with  the  work  to  be 
done,  lubrication  of  bearing,  kind  of  materials  used  for  the 
boxes,  and  diameter  of  journal. 

Line  shaft  bearings  are  designed  from  the  formula  : 


where    L  =  length  of  bearing, 

d=dia.  of  shaft  in  inches. 


SHOP  MATHEMATICS 


77 


The  projected  area  of  the  journal  is  the  size  used  in  the 
calculations  for  the  bearing  surface;  thus  the  area  of  bearing 
surface  for  a  4  inch  diameter  by  12  inch  journal  is  4 X  12  =  48 
square  inches. 

The  length  of  bearings  for  machine  spindles  vary  according 
to  the  service  required,  from  1£  to  6  times  the  diameter  of 
journal.  For  journals  under  2  inches  diameter  having  100 
to  1,000  R.  P.  M.  use  the  formula: 


The  allowable  pressure  on  the  ordinary  shaft  bearings  is 
40  pounds  per  square  inch  of  projected  area. 


BALL  BEARINGS 

Ball  bearings  are  used  especially  on  high  speed  and  light 
running  machinery.  The  diameter  of  the  ball  race  may  be 
calculated  from  the  size  and  number  of  balls  as  follows : 

In  Fig.  45  by  trigonometry 
AD 


let 


A0=__ 

sin  AOD 
8XAO=--D 


2n         n 

where  n  =  number  of  balls  in  race 

D  =  dia.  of  circle  at  center 

of  balls 
d=dia.  of  ball 

then         AD  =  -^ 

<i 

and        D+d=dia.  of  race  ring 


78 


SHOP  MATHEMATICS 


Example.    Find  the  diameter  of  ball  race  for  15,  \  inch  dia. 
balls. 


Solution. 


AD  =  —  =  .250  inches. 

1*°-  =  12° 


then 


sin  12°  =  .20791 

6)K. 

A0  = 


:=  1.20244 


.20791 

D  =  l .20244 X2  =  2.405  inches. 
D+d  =  2.405 +  .5  =  2.905 

A  clearance  of  about  T^  inch  is  usually  added  to  the 
diameter  of  the  ball  race  to  allow  for  slight  oversize  in  the 
diameter  of  the  balls. 

A  modification  of  the  ball  bearing  shown  in  Fig.  45  is 
sometimes  made  in  order  to  reduce  the  number  of  balls  used 
in  the  race  ring. 

This  is  shown  in  Fig.  45A  where  fewer  balls  are  held  from 
contact  with  each  other  by  a  separator  ring  or  cage. 
Then  the  formula  above 
AD 


A0  = 


sin  AOD 


becomes  A0  = 


where 


sin  AOD 

s 
T    ~2 


2n         n 

then  r  =  radius  of  balls  and  d  =  2r 

n  =  number  of  balls 
s  =  clearance  between  two  balls 
D  =  2X  A0  =  diameter  of  circle  at  center  of  ball 


SHOP  MATHEMATICS  79 

By  transposition  four  statements  are  obtained  as  follows: 

1.  Given  the  diameter  and  number  of  balls  and  the 
clearance,  to  find  D. 

s 

D  -    T+~*  °r      D ^o- 

2        .    180P  sin- 
sin n 

n 

2.  Given  the  number  of  balls,  the   clearance  and  D,   to 
find  r. 

D  I  .     18CP\        s         ,     _  /    .    180°\ 

r=-—(  sin  I —  or  d  =  D  [  sin I — s 

2  \  n    )        2  \  n    ) 

3.  Given  the  diameter  and  number  of  balls  and  D,  to 
find  s. 

n/   .     18CP\      , 
s  =  D I  sin 1  — d 

\          n    / 

4.  Given  D,  the  diameter  of  balls  and  clearance,  to  find  n. 

„.  .    180P  .        2r+s 

Find     sin  —  from  — K— 
n  D 


n 

18QP 


PROBLEMS 

1.  Find  the  dia.  of  ball  race  for  12,  ^  in.  dia.  balls. 

2.  Find  the  dia.  of  ball  race  for  15,  \  in.  dia.  balls. 

3.  What  is  the  dia.  of  ball  race  for  3,  £  in.  dia.  balls? 

4.  What  is  the  dia.  of  roll  race  for  12,  1£  in.  dia.  rolls? 

5.  Find  dia.  of  ball  race  for  22,  ^  in.  dia.  balls. 

6.  What  is  the  dia.  of  ball  race  for  26,  ^  in.  dia.  balls? 

7.  What  is  the  length  of  box  for  2^  in.  dia.  shafting? 


80  SHOP  MATHEMATICS 

8.  What  is  the  length  of  box  for  l^f  in.  dia.  shaft? 

9.  Find  length  of  box  for  2Jf  in.  dia.  shafting. 

10.  Find  length  of  hanger  bearing  for  3^  in.  dia.  shaft. 

11.  What  is  the  dia.  of  shafting,  when  bearing  for  boxes 
is  12  in,  long? 

12.  What  length  of  shaft  bearing  will  be  required  for 
5^  in.  dia.  shaft? 

13.  Find  length  of  box  for  6^  in.  dia.  shafting. 

14.  Find  dia.  of  shaft  for  hanger  box  16  in.  long. 

15.  What  dia.  of  shaft  will  be  required  for  box  9£  in. 
long? 

16.  When  40  Ibs.  is  allowed  per  sq.  in.  on  bearings,  what 
is  the  length  of  box    for  4^    in.    dia.    shaft,   that   carries 
9  pulleys,  with  3^  in.  belts  at  55  Ibs.  per  in.  strain,  each 
pulley  weighing  35  Ibs.? 

Note.     Use   approximate   formula   in   finding   weight   of   shaft. 
Page  172. 

17.  What  length  of  bearing  will  be  required  for  a  3^ 
in.  dia.  shaft,  carrying  6  pulleys  for  3^  in.  belts  at  55  Ibs.  per 
in.  strain  and  weighing  35  Ibs.  each,  allowable  pressure  40 
Ibs.  per  sq.  in.? 

18.  How  many  12  in.  dia.  pulleys  weighing  40  Ibs.  each 
for  4  in.  belts  at  55  Ibs.  per  in.  strain,  can  be  put  on  3  in. 
shaft  with  allowable  pressure  40  Ibs.  per  sq.  in.? 

19.  A  jack  shaft  has  3  pulleys  each  weighing  1,500  Ibs. 
The  pulleys  carry  26  in.  double  belts  at  82 ^  Ibs.  strain. 
Find  length  of  3  boxes  used,  at  8  ft.  on  centers,  when  shaft 
is  5  in.  dia.  and  projects  6  in.  beyond  centers  of  end  boxes, 
with  75  Ibs.  per  sq.  in.  pressure  on  bearing. 

20.  What  is  the  length  of  crank  shaft  bearing  for  7^  in. 
dia.  of  journal  when  shaft  is  8  ft.  long,   fly  wheel  weighs  8 


SHOP  MATHEMATICS  81 

tons,  belt  26  in.  wide  with  82^  Ibs.  per  in.  strain  and  80  Ibs. 
is  allowable  pressure  per  sq.  in.  on  bearing? 

21.  Find  number  of  3  in.  belt  pulleys  55  Ibs.  strain,  each 
pulley  weighing  100  Ibs.,  that  can  be  placed  on  a  4  in.  dia. 
line  shaft,  8  ft.  between  the  centers  of  hangers. 

22.  What  lengths  of  boxes  will  be  required  for  4^  in. 
dia.  jack  shaft,  8  ft.  long,  to  carry  2  drive  pulleys,  weighing 
1,000  Ibs.  each  with  24  in.  single  belt  at  55  Ibs.  strain  per  in., 
allowable  pressure  being  65  Ibs.? 

23.  A  jack  shaft  with  5  in.  journals,  with  75  Ibs.  pressure 
per  sq.  in.  will  require  2  boxes  of  what  length,  when  the  total 
weight  and  pull  of  belt  is  15,000  Ibs.? 

24.  What  is  the  bearing  surface  required  for  a  3^  in.  dia. 
line  shaft  with  40  Ibs.  allowable  pressure,  when  there  is  a 
weight  including  shaft  of  2,000  Ibs.  on  each  bearing? 

25.  Find  the  dia.   of  race  ring  for  10  ^  in.   dia.   balls 
when  the  clearance  between  the  balls  is  \  inch. 

26.  What  is  the  clearance  between  8  £  in.  dia.  balls  when 
the  race  ring  is  1\  in.  diameter? 

27.  Find  the  dia.  of  the  ball  to  be  used  in  a  15  ball  race 
ring  when  the  clearance  is  .114  in.  and  the  circle  at  the  center 
of  the  balls  is  If  in.  diameter. 

28.  How  many  \  in.  dia.  balls  will  be  required  for  a  race 
ring  \\  in.  dia.  when  the  clearance  between  the  balls  is 
.1365  inch? 

29.  Find  D  when  n  =  8,  d  =  f  in.  and  s  =  .250  in. 

30.  Find  n  whenD  =  l^in.,  d=  f  in.  and  s=.  199  in. 

31.  Find  s  when  d=  ^in.,n  =  20  andD  =  4in. 

32.  Find  d  whenD=  1.82  in.,n  =  6  ands=.41  in. 

33.  Find  s  when  w  =  9,  d=£in.  andD=  Ifin. 

34.  Find   s  whenZ)=  l^in.,  d=  ^  in.  and  w=  11. 


82  SHOP  MATHEMATICS 

MACHINE  KEYS 

Machine  keys  are  short  bars  of  metal,  usually  either  square 
or  rectangular  in  cross  section,  used  to  prevent  wheels, 
pulleys,  cranks,  or  other  pieces  from  rotating  on  the  shaft. 

Proportion  of  Machine  Keys 

Let  d=  diameter  of  shaft 
6  =  breadth  of  key 

t  =  thickness  of  key,  then  6  =  ——  and  t  =  — 

4  3 

Feather  keys  are  used  so  that  a  piece  can  slide  along  a  shaft 
and  yet  have  a  positive  rotation  with  the  shaft.  The  key 
is  usually  fastened  in  the  sliding  piece,  and  is  square  in  cross 
section  outline 

then         t=b—-j-. 
4 

PROBLEMS 

1.  A  6  in.  dia.  shaft  is  to  be  coupled  to  a  5  in.  shaft 
with  flange  couplings.    What  are  the  sizes  of  keys  required 
for  each  flange? 

2.  What  is  the  thickness  of  key  required  for  fastening 
pulley  to  a  2  in.  dia.  shaft? 

3.  What  size  of  key  will  be  required  to  fasten  a  24  in. 
crank  to  an  8  in.  dia.  shaft? 

4.  What  size  key  should  be  used  on  feed  worm  of  an 
engine  lathe,  when  feed  rod  is  1  in.  diameter? 

LINEAR  MEASURING  INSTRUMENTS 

The  meter  is  the  only  unit  for  linear  measure  legalized  by 
the  United  States  government,  but  the  bronze  bar  No.  11, 
given  to  the  United  States  by  the  British  government,  has 
been  generally  accepted  as  the  standard  unit  for  length  of 


SHOP  MATHEMATICS  83 

the  yard  at  a  temperature  of  61.79°  F.  From  this  bar  the 
makers  of  measuring  instruments  have  compared  their 
standards,  and  have  fixed  lengths  of  bars,  from  which  their 
instruments  are  made  with  as  fine  a  degree  of  accuracy  as  it 
is  possible  for  a  trained  and  skilled  mechanic  to  attain. 

When  it  is  remembered  that  the  bronze  No.  11  changes 
in  length  over  -ruthnr  m°h  f°r  each  degree  Fahrenheit,  it 
will  be  seen  that  extreme  accuracy  and  close  measurements 
depend  not  only  upon  the  accuracy  with  which  the  standard 
bar  is  made,  but  also  upon  the  temperature  of  the  bar  at 
the  time  that  measurements  are  taken  or  comparisons  made. 
Usually  each  shop  has  some  standard  bar,  which  is  referred 
to  as  a  standard  gauge,  with  which  all  other  measuring 
instruments  are  compared. 

The  micrometer  is  the  measuring  instrument  that  is  now 
generally  used  to  record  dimensions  of  any  body  in  thous- 
andths of  an  inch.  The  principle  of  construction  and  opera- 
tion of  the  micrometer  is  comparatively  simple.  It  consists 
of  a  screw  C,  Fig.  46,  having  forty  threads  per  inch,  that 
turns  through  a  stationary  nut  B,  which  is  part  of  a  curved 
bar  to  which  is  attached  a  round  pivot  E  called  the  anvil 
located  in  line  with  the  screw  C.  F  is  a  lock  nut  which  sets 
the  screw  C  firmly  in  any  desired  position  for  taking  dimen- 
sions. The  barrel  or  thimble  A  is  a  part  of,  or  fast  to  the 
screw  C,  and  is  divided  into  twenty-five  divisions  on  the 
beveled  end  at  a,  each  five  divisions  being  marked  with 
figures  0 — 5 — 10 — 15 — 20.  A  cylindrical  part  of  the  nut  B 
has  a  line  parallel  to  the  axis  of  the  longitudinal  motion  of 
the  screw,  and  at  right  angles  to  this  line  are  short  lines 
which  represent  the  position  of  the  end  of  thimble  A  for  each 


SHOP  MATHEMATICS 


revolution  of  the  screw  C.     The  first  one  marked  0  shows  the 

position  of  screw  when  it  just 
touches  the  end  of  anvil  E,  and  at 
this  position  the  0  line  on  thimble 
A ,  should  coincide  with  this  0  point 
at  its  intersection  with  line  parallel 
to  axis  of  screw  C.  As  one  turn  of 
a  forty  pitch  screw  is  equal  to  ?V  of  1 
inch,  or  r^^  inch,  the  motion  of 
thimble  A  through  the  space  of  one 
of  the  twenty-five  divisions  on  end 
of  thimble  will  give  a  longitudinal 
movement  of  the  screw  C  of  one- 
thousandth  inch  Cn^  inch).  Each 
four  turns  of  the  screw  C  will  be 
TfribX^,  or  TV0A  inch,  these  points 
being  marked  1,2,3,  4,  etc.,  up  to 
9,  indicating  100,  200,  300,  etc., 
thousandths  of  an  inch  between  end 
of  screw  C  and  anvil  E,  according  to 
the  position  of  the  beveled  end  of  thimble.  Suppose  a 
dimension  of  .5625  inch  is  to  be  taken,  the  screw  C  is 
opened  by  revolving  A  .until  0  on  thimble  corresponds  with 
the  axial  line  at  500  thousandths.  As  each  turn  of  C  is  equal 
to  25  thousandths  two  turns  are  made  with  A  which  gives 
550  thousandths.  Now  turning  A  through  12  spaces  on 
thimble  will  give  562  thousandths,  and  5  tenths  or  the  half 
thousandth  will  be  obtained  when  the  axial  line  is  half  way 
between  the  twelfth  and  thirteenth  division  lines  on  the 
thimble.  In  this  way  .5625  inch  can  be  obtained.  Other  di- 
mensions are  obtained  in  like  manner,  getting  the  number 
of  thousandths  required  between  the  hundredths  by  adding 


SHOP  MATHEMATICS  85 

25,  50,  or  75  for  one,  two,  or  three  complete  turns  of  the 
screw,  and  then  turning  the  thimble  any  number  of  spaces 
required  between  the  complete  turns  to  make  up  the  re- 
quired number  of  thousandths  for  getting  the  dimension 
wanted. 

When  a  vernier  scale  is  made  a  part  of  the  graduations 
of  a  micrometer  caliper,  it  is  unnecessary  to  get  the  fraction 
of  a  thousandth  of  an  inch  by  the  judgment  of  setting  the 
line  of  axis  between  two  spacing  lines  on  thimble,  as  the 
setting  can  be  found  directly  by  the  vernier  reading. 

THE  VERNIER 

The  vernier  principle  is  used  on  the  slide  caliper  for  setting 
the  jaws  of  the  caliper  to  -nsW  inch  and  is  as  follows:  the 
bar  A,  Fig.  47,  which  has  solid  jaw  B  at  one  end  graduated 
the  whole  length  into  fortieths  of  an  inch;  on  the  sliding  jaw  C 
is  a  scale  D,  which  has  twenty-five  divisions  on  the  beveled 
edge  equal  in  length  to  twenty-four  of  the  ¥V  inch  spaces 
on  the  bar.  It  is  evident  that  the  spaces  on  the  scale  D  will 
be  smaller  than  the  spaces  on  the  bar  by  ^V  °f  a  space  on 
the  bar,  as  24  of  them  are  equal  to  25  spaces  on  the  scale. 
Now  TjV  of  ?V  incn  =  T^W  inch, 

The  great  difficulty  in  setting  the  vernier  slide  caliper  is 
to  determine  which  lines  of  the  scale  and  bar  coincide  as, 
where  in  this  case  the  variations  in  width  of  spaces  are 
actually  ToV  o  inch,  in  the  micrometer  the  space  that  repre- 
sents TcyViT  inch  is  usually  ^  inch  or  more  in  width  on 
the  thimble  of  the  screw. 

The  vernier  calipers  are  set  to  any  dimensions  as  follows: 
if  a  dimension  of  1.128  in.  is  to  be  taken  the  0  mark  on  scale 
D  is  moved  until  it  coincides  with  the  1  inch  mark  of  the  bar 


86 


SHOP  MATHEMATICS 


As  each  division  on  the  bar  A  is  ^  inch  or  T^ 

the  0  line  on  the  scale  is  moved 
until  it  coincides  with  the  fifth 
line  from  the  1  inch  line  on  bar 
A,  the  set  screw  H  is  then 
tightened,  fastening  slide  F 
securely  to  bar  A,  E  being  left 
loose  to  allow  sliding  j  aw  C  to 
move  freely  along  bar  A  by  the 
turning  of  thumb  nut  G,  which  is  revolved  until  the  third 
line  from  0  on  the  scale  is  exactly  in  line  with  the  eighth  line 
from  the  1  inch  line  on  bar  A.  This  gives  the  required 
dimension  of  1.128  inch.  Other  dimensions  are  taken  in  the 
same  way. 

The  vernier  scale  on  the  micrometer  used  to  get  the  tenths 
of  thousandths  of  an  inch  is  made  as  follows:  Fig.  48  shows 
the  developed  projection  of  the  barrel  of 
nut  B  of  Fig.  46;  to  the  left  of  the  axial 
line  are  eleven  lines,  also  drawn  parallel 
to  the  axis  line  of  the  screw,  forming  ten 
spaces  equal  to  nine  of  the  twenty-five 
divisions  on  the  thimble  of  screw  C,  each 
of  these  spaces  being  smaller  by  one- 
tenth  than  one  division  on  the  thimble. 
These  lines  are  so  placed  on  the  barrel 
of  the  nut  that  the  first  one  coincides 
with  a  line  on  the  thimble  when  any  line  of  the  thimble  is  in 
line  with  the  first  axial  line  b,  Fig.  48.  Now  if  a  dimension  of 
one-tenth  of  some  even  one-thousandth  inch  is  required  the 
second  parallel  line  on  the  barrel  will  be  brought  into  line  with 
the  next  division  line  on  thimble,  if  two-tenths  over  are  re- 
quired, the  thimble  will  be  turned  until  the  third  line  coincides 


SHOP  MATHEMATICS  87 

with  the  third  parallel  line  on  the  barrel,  and  so  on  to  obtain 
any  number  of  tenths  over  the  required  dimension  wanted. 

Suppose  the  reading  .7658  is  to  be  found  on  the  micrometer 
that  has  a  vernier  attachment.  The  micrometer  is  set  first 
to  the  dimension  .765  as  given  above.  The  reading  for  .7658 
will  be  obtained  when  the  line  8  on  the  vernier  scale  exactly 
coincides  with  a  line  on  the  thimble. 

The  principle  of  the  micrometer  screw  thread  is  used  on 
different  machines  to  gauge  the  depth  of  a  cut  to  thousandths 
of  an  inch.  The  method  of  measuring  the  movement  of  the 
screw  is  the  same  as  for  the  micrometer  and  the  formula  is 
as  follows: 

Let  x  =  end  wise  movement  of  the  screw  or  nut  for  the 
rotation  through  one  division  on  the  disc  or 
thimble. 

N  =  number  of  divisions  on  the  disc  or  thimble  that 
is  fastened  on  screw. 

L  =  lead  of  screw  thread. 


Then  z  =  Lx  -; 
thus  in  the  micrometer, 

x  =  ?V  X  ^  =  TffW  inch. 

Example.  If  the  feed  screw  of  a  universal  grinder  bed  is 
|  inch  lead  and  the  disc  fast  to  the  screw  has  125  divisions, 
how  far  will  movement  of  disc  through  one  division  move 
bed  of  grinder  up  to  the  wheel? 

Solution.     By  formula: 


88  SHOP  MATHEMATICS 

PROBLEMS 

1.  What  is  the  lead  required  on  the  screw  of  grinder 
bed  to  move  bed  T^V^  m->  when  the  disc  on  screw  is  turned 
through  an  arc  of  3  degrees? 

2.  How  far  will  a  screw  of  ^g  in.  lead  move  a  nut  when 
turned  through  T^  of  a  turn? 

3.  When  the  cross  slide  on  a  lathe  is  operated  with  a 
screw  of  £  in.  lead,  what  is  the  movement  of  slide  when  a  disc 
of  100  divisions  on  the  screw  is  moved  through  an  arc  of  one 
division? 

4.  Find  the  lead  of  feed  screw  on  a  slide  that  will  move 
the  slide  ^^  in.  when  a  disc  with  100  divisions  is  turned 
through  an  arc  of  one  division. 

5.  If  the  cross  slide  feed  screw  of  a  certain  machine  is 
^  in.  lead,  what  ratio  of  gears  on  an  auxiliary  crank  rod 
geared  to  screw  will  be  required  on  rod  and  screw  to  move 
the  slide  TtfW  m->  when   the   graduated    disc   having    100 
divisions  on  the  crank  moves  through  an  arc  of  one  division? 

6.  If  the  adjustable  knee  of  a  shaper  is  to  be  raised 
T^o(T   in.   by  turning  crank  through  arc  of  one  space  of  a 
disc  fast  to  crank  rod  and  having  100  divisions,  and  the  screw 
that  stands  at  right   angles  to  crank   rod  has  ^  in.  lead, 
what  will  be  the  ratio  between  the  number  of  teeth  in  the 
beveled  gears  running  in  mesh  on  crank  rod  and  screw? 

7.  If  the  screw  that  moves  the  bed  of  a  milling  machine 
is  £  in.  lead,  what  number  of  spaces  should  be  made  on  a 
disc   to   move  the  bed  -r^W  m-   when  the  screw  is  moved 
through  the  arc  of  one  space  on  the  disc  that  is  fast  to  screw? 

8.  If  the  cross  feed  screw  of  a  lathe  is  i  in.  lead,  how  far 
will  the  movement  of  one  space  on  a  disc  of  25  divisions  that 
is  fast  to  screw  move  the  cross  slide? 


SHOP  MATHEMATICS  89 

9.  The  screw  that  raises  the  knee  on  a  milling  machine 
is  j%  in.  lead,  and  is  turned  by  a  crank  rod  with  a  pair  of 
bevel  gears  on  rod  and  screw.  On  crank  rod  is  a  disc  having 
100  divisions.  What  ratio  of  gears  will  be  needed  to  raise  the 
knee  T^Vtr  m-  when  the  disc  fast  to  the  crank  is  turned 
through  an  arc  of  one  division? 

10.  The  cross  feed  screw  for  a  milling  machine  bed  is  20 
pitch.    Into  how  many  divisions  should  the  disc  that  is  fast 
to  the  screw  be  divided  to   move  the  bed   ToW  m-   when 
the  disc  is  moved  through  the  arc  of  one  division? 

11.  If  the  table  of  a  horizontal  boring  mill  is  moved  with 
an  8  pitch  single  thread  screw,  into  how  many  divisions 
should  a  disc  fast  to  screw  be  divided  to  give  ^<y  in.  move- 
ment to  the  table  when  the  screw  is  moved  through  the  arc 
of  one  division  on  the  disc? 


MACHINES 

The  various  machines  used  for  manufacturing  purposes 
are  generally  classified  as  follows : 

Lathes;  as,  engine,  turning,  speed,  grinding,  cutting  off, 
axle,  wheel,  screw  machine,  etc. 

Planers;  as,  simpers,  slotters,  surface  grinders,  etc. 

Milling  machines;  as,  power,  hand,  duplex,  universal, 
vertical,  profiles,  etc. 

Drills;  as,  vertical  or  horizontal  spindles,  chucking  and 
boring  machines,  etc. 

Punches;  as,  presses,  shears,  etc. 

Power  hammers;  as,  steam,  helve,  trip,  drop,  etc. 

There  are  numerous  special  machines  which  can  be  classed 
as  modifications  of  some  one  of  the  above  classes. 

LATHE 

By  far  the  most  important  and  the  one  which  embodies 
the  general  principles  of  the  other  classes  is  the  lathe.  The 
figure  on  page  91  shows  outline  of  a  12  inch  by  5  foot  engine 
lathe. 

The  size  of  a  lathe  is  stated  by  the  largest  diameter  it  will 
swing  on  centers  and  the  length  of  the  bed.  The  carriage 
is  moved  along  the  length  of  the  Vs  of  the  lathe  bed  parallel 
to  the  axis  of  the  driving  spindle ;  the  motion  is  imparted  to 
carriage  from  the  driving  spindle  either  by  the  feed  rod  11, 
through  a  belt  transmission  that  is  used  when  doing  plain 
turning,  or  by  the  lead  screw  10,  and  a  system  of  change 
gears  which  is  used  when  cutting  screw  threads. 


SHOP  MATHEMATICS 


91 


10 


12"  x  3'0"  ENGINE  LATHE 


1  Head  stock 

2  Foot  or  tail  stock 

3  Carriage 

4  Elevating  tool  rest 

5  Apron 

6  Face  plate 

7  Back  gears 

9  Stud  of  feed  spindle 

10  Lead  screw 

11  Feed  rod 

12  Live  center 

13  Dead  center 

14  and  14i  Change  gears 

15  Intermediate  gear 

16  Upper  belt  feed  cone 


17  Lower  belt  feed  cone 

18  Tool  post 

19  Set  over  screw  for  tail  stock 

20  Back  gear  lever 

21  Reverse  feed  lever  for  car- 

riage 

22  Screw  feed  nut  lever 

23  Ball  crank  handle  for  tail 

spindles 

24  Ball  crank  handle  for  ele- 

vating rest 

25  Ball  crank  handle  for  cross 

feed 

25l  Ball  crank  handle  for  car- 
riage feed 


92  SHOP  MATHEMATICS 

The  stepped  cone  pulley  on  driving  spindle  of  lathe  is 
belted  to  similar  cone  on  countershaft  except  that  the  steps 
are  in  reverse  order. 

Example.  When  a  countershaft  makes  160  R.  P.  M., 
what  is  the  R.  P.  M.  of  a  lathe  spindle,  with  pulley  7f  in.  dia. 
on  counter,  belted  to  a  4f  in.  dia.  on  lathe  spindle? 

Solution.     By  formula  for  speeds  of  pulleys, 


4% 

PROBLEMS 

1.  When  the  dia.  of  lathe  pulley  is  7f  in.  and  counter- 
shaft pulley  is  4f  ,  what  is  the  R.  P.  M.  of  spindle  if  counter 
makes  160  R.  P.  M.? 

2.  Find  R.  P.  M.  of  a  lathe  spindle  when  the  pulley  is  6£ 
in.  dia.  and  countershaft  makes  155  R.  P.  M.  with  a  pulley 
6  1  in.  diameter. 

3.  Find  R.  P.  M.  of  a  lathe  cone  pulley  8^  in.  dia.  when 
the  countershaft  makes  170  R.  P.  M.  and   the  pulley   on 
countershaft  is  5  in.  diameter. 

4.  When  a  lathe  cone  pulley  is  3J  in.  dia.  and  counter- 
shaft pulley  is  11^  in.  dia.,  find  R.  P.  M.  of  spindle  when 
countershaft  makes  165  R.  P.  M. 

6.  When  a  lathe  spindle  pulley  is  8^  in.  dia.  and  counter- 
shaft runs  170  R.  P.  M.  with  a  pulley  4£  in.  dia.,  what  is 
the  R.  P.  M.  of  the  lathe  spindle? 

6.  When  a  cone  pulley  is  3£  in.  dia.  and  countershaft 
runs  195  R.  P.  M.  with  pulley  9£  in.  dia.  on  countershaft, 
what  is  the  R.  P.  M.  of  the  lathe  spindle? 

7.  The  cone  pulley  of  a  certain  lathe  has  steps  7|  in., 
6£  in.,  4f  in.,  and  3£  in.  diameters  respectively,  belted  to 


SHOP  MATHEMATICS  93 

the  steps  of  a  cone  on  countershaft,  the  diameters  of  which 
are  as  follows:  4f  in.,  6£  in.,  7f  in.,  and  9i  in.  When  the 
countershaft  runs  170  R.  P.  M.  find  the  different  speeds  at 
which  the  work  on  the  spindle  may  be  driven. 

8.  A  lathe  has  on  its  driving  spindle  a  four  step  cone 
pulley  with  diameters  as  follows:   8J  in.,  6f  in.,  5  in.,  and 
3i  in.    Belted  to  this  cone  is  the  cone  on  countershaft  with 
steps  5j  in.,  6f  in.,  8^  in.,  and  10J  in.  diameters  respective!}'. 
Find  the  different  R.  P.  M.  at  which  work  can  be  turned 
in  the  lathe,  when  the  countershaft  runs  180  R.  P.  M. 

9.  A  lathe  has  a  cone  on  the  head  spindle  with  steps  8^ 
in.,  6f  in.,  5^  in.,  and  3£  in.  diameters  respectively.    If  the 
countershaft  runs  175  R.  P.  M.  with  a  cone  having  steps 
4|  in.,  6j  in.,  7|  in.,  and  9^  in.  diameters  respectively,  what 
different  R.  P.  M.  may  be  given  the  work  on  the  lathe  cen- 
ters? 

BACK  GEARS 

The  back  gears  are  used  to  reduce  the  speed  of  lathe  spindle, 
thus  allowing  a  heavier  cut  to  be  taken  at  reduced  speed. 

On  the  back  gear  spindle  are  two  gears  fast  to  the  same 
sleeve;  on  the  lathe  cone  pulley  is  fastened  a  small  gear 
which  may  be  thrown  into  mesh  with  large  gear  on  back  gear 
sleeve ;  on  the  other  end  of  the  cone  pulley  fastened  securely 
to  lathe  spindle  is  a  large  gear  which  is  thrown  into  mesh 
with  the  smaller  gear  of  back  gears.  When  the  back  gears 
are  not  in  mesh,  a  clamp  nut  fastens  spindle  gear  securely 
to  cone  pulley. 

Example.  If  a  countershaft  running  160  R.  P.  M.  has 
a  5  in.  dia.  pulley  belted  to  8  in.  pulley  on  lathe  spindle, 
with  gear  fast  to  cone  having  32  teeth  and  gear  fast  to  spindle 
having  90  teeth,  these  being  in  mesh  with  back  gears  of  88 
and  30  teeth  respectively,  what  is  the  R.  P.  M.  of  lathe  driv- 
ing spindle? 


94  SHOP  MATHEMATICS 

Solution.     By  formula  for  driver  and  follower, 
20        4 


33-  .  P.M. 

11    3 

PROBLEMS 

1.  When  a  countershaft  runs  165  R.  P.  M.  and  pulley  on 
countershaft  is  4f  in.  dia.  belted  to  a  7£  in.  dia.  pulley  on 
lathe  spindle;   and  when  the  gears  are  the  same  as  in  above 
example,  what  is  the  R.  P.  M.  of  lathe  spindle? 

2.  When  the  pulley  on  a  countershaft  is  7f  in.  dia.  and 
the  cone  pulley  is  4f  in.  dia.  other  sizes  same  as  in  problem  1 
find  R.  P.  M.  of  spindle. 

3.  When  a  countershaft  makes  155  R.  P.  M.  and  the 
pulley  is  8^  in.  dia.  belted  to  lathe  cone  pulley  6f  in.  dia. 
with  back  gears  85  and  32  teeth,  in  mesh  with  31  on  cone 
and  84  on  spindle,  what  is  the  number  of  R.  P.  M.  of  lathe 
spindle? 

4.  When  a  countershaft  runs  185  R.  P.  M.  and  the  pulley 
on  countershaft  is  5  in.  dia.  belted  to  an  8£  in.  dia.  pulley 
on  spindle  with  gears  same  as  in  problem  3,  what  is  the 
number  of  R.  P.  M.  of  lathe  spindle? 

5.  When  a  countershaft  pulley  6  in.  dia.  running  160 
R.  P.  M.  is  belted  to  a  cone  pulley  6f  in.  dia.  and  back  gears 
with  75  and  20  teeth  in  mesh  with  25  tooth  gear  on  cone  and 
69  tooth  gear  on  spindle,  what  is  the  number  of  R.  P.  M. 
of  spindle? 

6.  When  a  countershaft  pulley  5£  in.  dia.  and  running 
225  R.  P.  M.  is  belted  to  a  cone  pulley  on  the  lathe  spindle 
7£  in.  dia.  with  gears  on  back  gear  quill  of  75  and  25  teeth 
in  mesh  with  25  teeth  on  cone  pulley  and  65  teeth  on  spindle, 
what  is  the  R.  P.  M.  that  a  piece  will  make  on  lathe  centers? 


SHOP  MATHEMATICS  95 

SCREW  THREAD  CUTTING 

One  of  the  most  important  operations  on  the  lathe  is  the 
cutting  of  screw  threads  with  a  tool  formed  to  make  the  exact 
shape  of  thread.  In  order  to  cut  the  required  number  of 
threads  per  inch,  the  work  must  revolve  an  exact  number  of 
times,  equal  to  the  number  of  threads  to  be  cut,  while  the 
carriage  which  holds  the  thread  cutting  tool  firmly  during 
the  cut  is  moved  along  the  bed  of  lathe  exactly  one  inch. 
Now  as  the  carriage  feed  is  liable  to  variable  motion  with  the 
belt  feed  owing  to  slip  of  the  belt  a  positive  uniform  motion 
is  given  to  the  cutting  tool  by  a  system  of  change  gears  and 
lead  screw  arranged  to  give  any  desired  feed  of  carriage  for  the 
required  number  of  revolutions  of  lathe  spindle,  or  work  to 
be  cut. 

The  illustration  on  page  91  shows  details  of  lathe:  10  is  the 
lead  screw  and  14  is  the  gear  that  imparts  a  positive  motion  to 
10  by  means  of  a  feather  key.  This  gear  is  held  in  place  with  a 
nut  on  the  end  of  lead  screw;  9  is  the  change  gear  spindle 
on  which  any  gear  may  be  fastened  with  a  feather  key  and 
nut.  Inside  of  the  head  casting  1  is  a  gear  in  mesh  with  the 
gear  on  the  driving  spindle  of  the  lathe.  If  these  two  gears 
are  in  the  ratio  of  1  to  1,  then  when  the  work  makes  one 
revolution  9  will  also  make  one  revolution  imparting  motion 
to  lead  screw  10  with  the  velocity  ratio  equal  to  the  ratio 
of  the  number  of  teeth  in  change  gears  14  and  14^  For  ex- 
ample, suppose  lead  screw  10  has  5  threads  per  inch,  or  £ 
inch  lead,  and  a  screw  of  the  same  number  of  threads  is  to 
be  cut  on  the  lathe  centers,  what  gears  will  be  required  on 
stud  9  and  end  of  lead  screw  10? 

Solution.  Drive  spindle  must  revolve  five  times  while  lathe 
carriage  moves  along  the  bed  1  inch.  As  lead  screw  10, 


96  SHOP  MATHEMATICS 

which  gives  it  a  positive  motion  is  5.  pitch  single  thread,  it 
must  revolve  five  times  to  give  1  inch  motion  to  carriage; 
therefore  9  and  10  must  each  make  five  turns  in  the  same 
time  and  gears  14  and  14:  will  be  the  same  size,  or  will  have 
a  velocity  ratio  of  1  to  1.  Therefore  any  two  gears  having 
the  same  number  of  teeth  as  25  and  25  may  be  used. 

A  general  formula  for  change  gears  is  as  follows : 

Pitch  of  lead  screw  No  of  teeth  of  stud  gear 

Pitch  of  screw  to  be  cut     No.  of  teeth  of  lead  screw  gear 

Example.  If  the  pitch  of  the  lead  screw  is  5  and  it  is  re- 
quired to  cut  a  10  pitch  screw. 

Then  T5<y  =  i  or  gear  on  lead  screw  must  have  double 
the  number  of  teeth  of  the  gear  on  the  stud.  Then  any  two 
gears  which  have  this  ratio  may  be  used. 

The  abov£  formula  may  also  be  stated  in  the  form  of  a  rule 
as  follows: 

RULE.  The  product  of  number  of  threads  to  be  cut,  by 
the  number  of  teeth  in  spindle  (or  stud)  gear,  is  equal  to 
the  product  of  the  number  of  teeth  in  the  gear  on  lead  screw 
by  the  number  of  threads  in  the  lead  screw. 

If  the  gears  on  stud  9  and  driving  spindle  of  the  lathe  are 
not  in  the  ratio  of  1  to  1  the  ratio  can  be  found  by  the  follow- 
ing method:  putting  two  gears  of  the  same  size  on  9  and  10, 
cut  a  trial  piece  and  count  the  threads  per  inch.  If  on  trial 
it  is  found  that  the  number  of  threads  cut  per  inch  is  twice 
the  number  per  inch  of  the  lead  screw  then  the  velocity 
ratio  of  the  spindle  and  stud  is  2  to  1.  Then  in  calculating 
the  ratio  of  the  change  gears  required,  the  above  formula 
could  be  changed  to  read  as  follows: 

Pitch  of  lead  screw  X  2  No.  of  teeth  in  stud  gear 

Pitch  of  screw  to  be  cut     No.  of  teeth  in  lead  screw  gear 


SHOP  MATHEMATICS  97 

If  the  velocity  ratio  is  not  given,  1  to  1  will  be  understood 
in  the  problems  for  screw  cutting  on  the  lathe. 

PROBLEMS 

1.  What  change  gears  can  be  used  to  cut  an  11^  pi. 
thread,  when  lead  screw  is  5  pitch? 

2.  What   change   gears   can   be   used  to  cut  a  12  pi. 
thread,  when  lead  screw  is  5  pi.  and  stud  gear  has  25  teeth? 

3.  Find  the  stud  gear  to  use  to  cut  16  threads  per  in. 
when  lead  screw  is  5  pi.  and  screw  gear  has  80  teeth. 

4.  Find  the  screw  gear  to  cut  18  pi.  screw  thread  when 
lead  screw  is  6  pi.  and  stud  gear  has  24  teeth. 

5.  When  the  velocity  ratio  of  driving  spindle  and  stud 
is  2  to  1  and  lead  screw  is  8  pi.,  what  is  the  number  of  teeth 
in  screw  gear  to  cut  1 1  threads  per  inch  when  stud  gear  has 
96  teeth? 

Note.  The  gears  on  inside  end  of  spindle  are  arranged  on  a 
pivoted  lever  so  that  a  train  of  three  gears  shall  be  in  mesh,  or  by 
throwing  over  lever  21  a  train  of  four  gears  can  be  put  into  mesh, 
thus  giving  a  feed  motion  of  carriage  in  either  direction  according 
as  lever  is  thrown  up  or  down;  this  mechanism  is  called  the  rever- 
sing feed  gears. 

6.  When  the  reversing  feed  gears  are  in  the  ratio  of 
2  to  1  with  driving  spindle,  what  stud  gear  will  be  required 
to  cut  16  threads  per  inch,  if  lead  screw  is  8  pi.  and  screw 
gear  has  48  teeth? 

7.  What  stud  gear  will  be  used  to  cut  a  9  pi.  thread, 
when  lead  screw  is  6  pi.  and  the  screw  gear  has  36  teeth? 

8.  What  screw  gear  will  be  used  to  cut  a  16  pi.  thread, 
when  lead  screw  is  6  pi.  and  stud  gear  has  24  teeth? 

9.  What  change  gears  can  be  used  to  cut  a  5  pi.  thread 
when  lead  screw  is  6  pitch? 


98  SHOP  MATHEMATICS 

10.  What  change  gears  can  be  used  to  cut  36  pi.  threads, 
when  lead  screw  is  6  pi.  and  the  smallest  gear  of  the  set  has 
24  teeth? 

COMPOUND  GEARING 

To  save  making  so  many  change  gears  in  a  set  for  thread 
cutting  and  especially  to  avoid  using  gears  with  a  large  num- 
ber of  teeth,  a  pivot  is  placed  between  the  stud  and  inter- 
mediate gear  15,  on  which  revolves  two  gears  fast  together 
called  a  compound  gear.  These  gears  are  usually  in  the  ratio 
of  2  to  1  but  the  principle  of  operation  would  be  the  same  if 
gears  in  any  other  ratio  were  used.  Thus  the  principle  of 
compound  gearing  comes  under  the  rule  for  drivers  and  fol- 
lowers, page  12,  where  another  driver  and  follower  has  been 
brought  into  the  calculations. 

When  the  velocity  ratio  is  2  to  1  the  number  of  teeth  in 
followers  can  be  multiplied  by  2  and  the  drivers  by  1  rather 
than  to  count  up  the  number  of  teeth  in  the  compound  gears. 

Example.  With  a  2  to  1  compound  gear  having  24  teeth 
and  lead  screw  6  pi.,  what  gear  will  be  required  on  lead  screw 
to  cut  36  threads  per  inch? 

Solution.     With  simple  change  gears  the  proportion  will 

*p    .--.7)  be  ~^  =  —  •'•  x  =  144,  which  makes  a 

/  V    .  V-'-  gear  of  a  large  number  of  teeth. 

'•/'• -/*•''        \  With  a  compound  gear  arranged 

'-.,  ';        ~      as  in  Fig.  49,  the  proportion  will  be 

/'"'•<  ,  „          6X2    24  . 

..".   °   :         as  follows  :-—--  =  —  ..x  =72  teeth 

"  *  • " "        '•         /  OO  X 


m  lead  screw  gear.     By  throwing 
stud  gear  out  of  mesh  an  18  pi.  thread 
can  be  cut.     The  intermediate  gear  M   serves  the  purpose 


SHOP  MATHEMATICS  99 

of  a  belt  except  that  it  imparts  a  positive  motion,  therefore 
it  is  not  brought  into  the  calculations. 

D  =  the  gear  on  stud  that  meshes  with  lathe  spindle  through 
the  reversing  gear  train. 

L  =  the  large  gear  of  the  compound. 

C  =  the  small  gear  of  the  compound. 

S  =  gear  on  end  of  lead  screw. 

D  and  C  are  the  drivers. 

L  and  S  are  the  followers. 

It  frequently  happens  that  the  velocity  ratio  between  the 
lead  screw  and  the  head  spindle  is  such  that  no  two  gears 
of  the  set  furnished  with  the  lathe  have  the  required  ratio. 
In  such  cases,  two  sets  of  gears  may  be  used,  the  product  of 
whose  ratios  gives  the  ratio  required. 

Example.  Find  the  gearing  required  to  cut  a  30  pitch 
thread  on  a  lathe  having  a  5  pitch  lead  screw. 

5111 

Solution.     The  ratio  —  -^=-^-  =  -^-X-^-'    Therefore  one  set 
oO      o       o       Si 

of  gears  in  the  ratio  of  1  to  3  and  a  second  set  in  the  ratio  of 
1  to  2  would  be  needed;  and  the  gears  24,  48,  30  and  90 
would  be  correct,  24  and  30  being  the  drivers  and  48  and  90 
the  followers. 

Example.  Find  the  compound  gearing  required  to  cut  a 
35  pitch  thread  on  a  lathe  whose  lead  screw  is  6  pitch. 

J> 3X3    3X10    2X20    30     J,O_ 

35  =  7X5  =  7X10  X  5X20  ~  70    WO 

Therefore  gears  30  and  40  will  be  used  as  drivers  and  70 
and  100  as  followers. 


100  SHOP  MATHEMATICS 

These  principles  are  summarized  in  the  following: 
RULE  FOR  COMPOUND  GEARING. 

1.  Find  the  ratio  between  the  pitch  of  the  lead  screw  and 
the  pitch  of  the  screw  to  be  cut. 

2.  Separate  each  term  of  this  ratio  into  two  factors  and 
express  the  result  as  the  product  of  two  fractions. 

3.  Multiply  both  numerator  and  denominator  of  each 
fraction  by  such  a  number  as  will  make  each  product  repre- 
sent the  number  of  teeth  found  hi  one  of  the  gears  furnished 
with  the  lathe.    The  numerators  of  these  two  fractions  will 
denote  the  driving  gears,  and  the  denominators  the  follower 
gears. 

PROBLEMS  IN  COMPOUND  GEARING 

1.  The  gears  furnished  with  a  14  in.  Reed  lathe  have 
24,  32,  40,  44,  48,  52,  56,  60,  64,  72,  and  110  teeth,  and  the 
lead  screw  is  6  pitch.     Find  the  compound  gears  required 
for  cutting  a  3^  pitch  thread. 

2.  With  the  same  set  of  gears  and  same  pitch  lead  screw 
as  given  in  problem  1,  find  compound  gearing  required  for 
cutting  a  25  pitch  screw.     A  28  pitch  screw.     A  7%  pitch 
screw.     A  screw  of  f  in.  lead. 

3.  The  change  gears  furnished  with  a  13  in.  Reed  lathe 
have  25,  30,  35,  40,  45,  50,  55,  60,  65,  69,  70,  80,  90,  100, 
110  and  120  teeth,  and  lead  screw  has    5    threads  to  the 
inch.     Find  the  compound  gearing  required  for  cutting  an 
1 1  £  pitch  pipe  thread. 

4.  With  the  same  lathe  and  same  set  of  gears,  find  the 
compound  gearing  required  for  cutting  a  40  pitch  screw. 
A  23  pitch  screw.    A  27  pitch  screw.     A  screw  of  ^  inch 
lead.    A  screw  of  ^  inch  lead. 


SHOP  MATHEMATICS  101 

TURNING  TAPERS 

Lathes  are  also  used  to  turn  tapers  on  round  stock,  and 
although  there  are  several  attachments  in  use  for  taper 
turning,  a  common  method  is  that  of  offsetting  the  tail  or 
foot  stock  center  so  that  it  is  out  of  line  with  driving  center 
the  necessary  amount  to  produce  the  required  taper. 

When  the  taper  is  to  be  turned  the  whole  length  of  the  bar, 
the  dead  center  is  offset  one-half  the  difference  between  the 
diameters  at  each  end  of  piece.    The  formula  is, 
_ 


x  =  offset  of  center. 
D  =  diameter  at  large  end  of  taper. 
d  =  diameter  at  small  end  of  taper. 

When  the  taper  runs  only  a  part  of  the  length  of  piece,  the 
amount  of  offset  is  determined  by  the  length  of  piece  between 
the  centers,  and  the  diameters  at  each  end  of  the  tapered 
part.  The  formula  becomes 


x  =  L  ( 


I 

where     x  =  offset  of  center. 

D  =  diameter  at  large  end  of  taper. 
d  =  diameter  at  small  end  of  taper. 
L  =  whole  length  of  piece. 
1  =  length  of  the  part  tapered. 

Example.     What  is  the  offset  of  dead  center,  when  a  bar 
36  inches  long  is  to  be  turned  9  inches  of  its  length,  tapered 
from  2  inch  dia.  at  end  to  4  inch  diameter? 
Solution.     Bv  formula 


102  SHOP  MATHEMATICS 

The  taper  per  inch  in  a  turned  piece  is  found  by  subtracting 
the  small  diameter  from  the  large  and  dividing  the  remainder 
by  the  length;  as  a  piece  8  inches  long  with  diameters  at 
ends  1  and  1\  inches  is  tapered  ^  inch  per  inch  in 

length  or  Y  =  ^  =  J-  =  iXi  =  *V 

When  it  is  desired  to  find  the  taper  per  foot,  the  formula  is 

=  12(D—  d) 
IJ~         I 
where    y  =  the  taper  per  foot  and  other  values  same  as  given 

above. 

Example.  What  is  the  taper  per  ft.  in  the  example  given 
above? 

Solution.     Bv  formula: 

which  s.  the 


L  tJ  «7 

taper  per  foot. 

PROBLEMS 

1.  A  bar  of  steel  is  tapered    1^  in.  per  ft.   and  diam- 
eters at  end  of  taper  are  1  in.  and  If  in.     Find  length  of 
taper. 

2.  A  bar  is  tapered  16  in.  of  its  length  and  diameters 
at  ends  of  taper  are  1  in.  and  2^  in.    Find  taper  per  foot. 

3.  A  bar  of  steel  is  tapered  7  in.  of  its  length,  the  diam- 
eters at  ends  of  taper  are  2J  in.  and  3f  in.     What  is  the 
taper  per  foot? 

4.  A  bar  is  turned  on  a  taper  24  in.  of  its  length.    The 
diameters  at  each  end  of  taper  are  f  in.  and  ^  in.     What 
is  the  taper  per  foot? 

5.  A  shank  mill  has  a  tapered  end  5^  in.  long.     The 
taper  is  ^  in.  and  ^f  in.  diameters  at  the  ends.     What  is 
the  taper  per  foot? 


SHOP  MATHEMATICS  103 

6.  A  small  cutter  has  shank  tapered  3  in.  long,  the  dia. 
at  small  end  is  .365  in.,  at  large  end  the  taper  is  .525  in.  dia. 
What  is  the  taper  per  foot? 

7.  The  taper  shank  on  a  cutter  is  3^  in.  long,  the  small 
dia.  is  .573  in.,  and  the  large  dia.  is  .749  in.     What  is  the 
taper  per  foot? 

8.  The  taper  on  a  reamer  is  8^  in.  long,  the  small  dia.  is 
2^  in.,  the  large  dia.,  is  2^  in.    What  is  the  taper  per  foot? 

9.  Find  the  taper  per  ft.  of  a  pin  2£  in.  long,  and  £  in. 
and  f  in.  diameters  respectively  at  ends. 

10.  What  is  the  offset  of  dead  center  for  turning  a  taper 

16  in.  long  on  the  end  of  bar  40  in.  long,  when  the  diameters 
at  ends  of  the  taper  are  1J  in.  and  4  inches? 

11.  What  is  the  offset  of  center  for  turning  a  taper  19  in. 
long  on  a  bar  25  in.  long,  if  the  diameters  at  ends  of  taper 
are  1  in.  and  If  inches? 

12.  A  bar  is  39  in.  long;  a  taper  11  in.  long  turned  on  one 
end  is  2|  in.  dia.  at  small  end  and  3j%  in.  dia.  at  large  end. 
What  is  the  offset  of  the  center  for  turning  the  taper? 

13.  Find  the  offset  of  the  center  for  turning  a  taper  30  in. 
long  on  a  bar  34  in.  long,  when  the  diameters  of  taper  are  \\ 
in.  and  If  inches. 

14.  Find  the  offset  of  the  center  for  turning  a  taper  33  in. 
long  on  a  bar  44  in.  long,  with  the  diameters  3£  in.  and  4\  in. 
on  ends  of  taper. 

15.  What  is  the  offset  for  a  lathe  center  to  turn  a  taper 

17  in.  on  a  bar  19  in.  long,  if  the  diameters  of  taper  are  1£  in. 
and  1 1  in.  on  ends? 

16.  How  much  should  the  dead  center  be  moved  over  to 
turn  u  taper  bar  4  in.  long,  when  one  end  of  piece  is  .779  in. 
dia.  and  the  other  end  is  .982  in.  diameter? 


104  SHOP  MATHEMATICS 

17.     What  is  the  taper  per  ft.  in  problem  16? 

Note.  The  formula  for  offset  of  lathe  center  for  turning  tapers, 
is  very  helpful.  It  is,  however,  only  a  close  approximation  as  the 
calculation  is  only  exact  between  the  ends  of  centers,  whereas,  the 
stock  to  be  turned  is  supported  a  little  way  up  the  countersunk 
part  on  each  end. 

CUTTING  SPEEDS 

The  speed  at  which  any  piece  of  work  should  revolve  in 
taking  a  cut  on  the  lathe  depends  on  the  following  conditions: 

(1)  The  kind  and  condition  of  the  material  of  which  the 
work  is  made. 

(2}     The  kind  and  condition  of  the  cutting  tool. 

(3)  The  lubrication  of  the  tool  while  making  the  cut. 

(4)  The  size  of  the  chip. 

Therefore  any  rule  for  speeds  will  be  only  approximate, 
but  from  experience  with  the  carbon  steels  which  are  used 
most  in  the  toolmaking  department  for  the  finishing  cuts,  the 
following  speeds  have  been  found  practical  as  a  basis  for 
trial: 

Soft  brass  80  F.  P.  M. 

Gray  iron  castings          Jfi  F.  P.  M. 

Machinery  steel  30  F.  P.  M. 

Annealed  tool  steel        20  F.  P.  M. 

Note.  The  high  speed  steel  cutting  tools  can  be  used  at  about 
double  the  above  speed. 

The  F.  P.  M.  can  be  found  for  lathe  turning  from  the 
formula : 

irRd 

=  12 

C=  cutting  speed  in  F.  P.  M. 
R=R.P.  M.  of  the  work. 
d=  diameter  of  work  in  inches. 


SHOP  MATHEMATICS  105 

The  approximate  rule  for  finding  the  F.  P.  M.  is  to  count 
the  revolutions  for  a  quarter  of  a  minute  and  multiply  this 
number  by  the  diameter  of  the  work  in  inches.  This  rule 
is  based  on  the  fact  that  if  in  formula  just  given  d=l  inch, 

then  -—  =  —  foot  approximately. 

•*.#  "T 

Then  approximately, 
Rd 

"T 

and  R^~ 


PROBLEMS 

In  the  following  problems  the  approximate  formula  for  R 
will  be  used  unless  otherwise  stated. 

1.  A  bar  of  4  in.  dia.  cold  rolled  machinery  steel  is  to 
be  turned  in  lathe.    What  approximate  R.  P.  M.  should  it 
revolve  to  turn  a  chip  at  30  F.  P.  M.? 

2.  What  is  the  approximate  R.  P.  M.  for  turning  a  chip 
on  a  tool  steel  reamer  1  in.  diameter? 

3.  A  cast  iron  pulley  10  in.  dia.  is  to  have  the  rim  face 
turned  in  the  lathe.    How  many  R.  P.  M.  should  it  make? 

4.  How  many  R.  P.  M.  should  a  brass  rod  f  in.  dia.  make 
in  taking  a  cut  in  the  lathe? 

5.  The  cast  iron  cylinder  of  a  gasolene  motor  5  in.  dia. 
is  to  be  bored  in  lathe.    How  many  R.  P.  M.  should  a  boring 
bar  make  to  do  the  work? 

6.  If  a  ring  gauge  of  tool  steel  is  to  be  bored  in  a  chuck 
to  2  in.  dia.,  what  is  the  approximate  R.  P.  M.  that  it  should 
make? 


106  SHOP  MATHEMATICS 

7.  The  hub  of  a  cast  iron  pulley  is  to  be  bored  for  finish- 
ing with  a  3^  in.  dia.  reamer.    How  many  R.  P.  M.  should 
it  make  approximately  in  the  chuck? 

8.  A  brass  nut  is  to  be  threaded  in  the  lathe  for  a  \\  in. 
bolt.    What  would  be  the  maximum  speed  in  R.  P.  M.? 

9.  A  machinery  steel  collar  is  to  be  shrunk  onto  a  H 
in.  arbor.    What  speed  should  it  revolve  when  boring  to  size? 

10.  A  crank  pin  4£  in.  dia.  is  to  be  shrunk  into  place. 
AVhat  approximate  R.  P.  M.  should  it  make  in  turning  to 
size,  if  made  of  annealed  tool  steel? 

11.  A  cast  iron  pulley  16  in.  dia.  should  turn  approxi- 
mately at  what  R.  P.  M.  for  turning  the  rim? 

12.  A  l^f  in.  cold  rolled  machinery  steel  shaft  is  to  be 
turned  one  chip  in  depth  near  the  end  to  fit  a  flange  coupling. 
How  many  R.  P.  M.  should  it  make? 

13.  A  crank  shaft  made  of  cold  rolled  machinery  steel 
is  to  have  the  bearings  finished  7}  in.  dia.     What  speed  in 
R.  P.  M.  should  it  have? 

MISCELLANEOUS  PROBLEMS 

14.  When  a  main  line  of  shafting  runs  at  200  R.  P.  M. 
with  a  10  in.  dia.  driver  belted  with  a  14  in.  pulley  on  coun- 
tershaft, what  R.  P.  M.  will  a  lathe  spindle  make  that  has  a 
9  in.  dia.  cone  pulley  belted  to  a  6  in.  pulley  on  counter? 

15.  When  main  line  shafting  runs  150  R.  P.  M.  with  a  16 
in.  driver  belted  to  a  12  in.  follower  on   countershaft,  how 
many  R.  P.  M.  will  the  spindle  make  with  a  3  in.  dia.  pulley 
belted  to  a  7  in.  pulley  on  counter? 

16.  When  a  main  line  runs  150  R.  P.  M.  from  a  14  in. 
driver  belted  to  10  in.  dia.  pulley  on  counter,  cone  pulley  on 


SHOP  MATHEMATICS  107 

counter  8^  in.  belted  to  6J  in.  dia.  on  lathe  spindle,  find 
R.  P.  M.  of  spindle  when  gear  fast  to  cone  has  28  teeth  in 
mesh  with  88  teeth  on  back  gear  and  36  teeth  on  back  gear 
in  mesh  with  96  teeth  gear  fast  to  spindle? 

17.  If  2%  is  allowed  for  slip  on  each  belt  in  problem  16, 
what  R.  P.  M.  will  spindle  make? 

18.  What  dia.  of  pulley  should  be  put  on  a  countershaft 
to  make  175  R.  P.  M.  when  the  pulley  is  to  be  belted  to  a 
15  in.  dia.  on  main  line  making  160  R.  P.  M.? 

19.  A  lathe  has  cone  pulley  9  in.  dia.  belted  to  a  counter- 
shaft cone  12  in.  dia.  with  back  gears  39  and  88  teeth  in 
mesh  with  65  teeth  on  spindle  and  27  teeth  on  cone;    the 
lathe  is  to  bore  a  4£  in.  dia.  cast  iron  disc  at  40  F.  P.  M. 
When  the  dia.  of  drive  pulley  on  main  line  is  15  inches, 
what  dia.  of  follower  pulley  on  the  countershaft  will  be  re- 
quired if  main  line  runs  150  R.  P.  M .? 

20.  An  axle  lathe  is  to  turn  4^  in.  dia.  crucible  steel  axles 
at  9  F.  P.  M.;  the  back  gears  have  66  and  22  teeth  in  mesh 
with  26   and  58  teeth  on  cone  and  spindle  respectively. 
When  main  line  runs  160  R.  P.  M.  and  dia.  of  driver  on  main 
line  is  10  in.,  what  dia.  of  follower  on  counter  will  be  required, 
if  a  12  in.  dia.  pulley  on  lathe  spindle  is  belted  to  11  in.  dia. 
pulley  on  countershaft? 

21.  If  2%  were  to  be  allowed  for  slip  in  each  of  the  two 
belts  of  problem  20,  how  much  larger  dia.  of  drive  pulley  will 
be  required  on  main  line? 

22.  A  crank  pin  is  to  have  a  taper  seat  in  crank  4^  in. 
long.    The  diameters  at  each  end  of  taper  are  4£  in.  and  4^ 
in.    What  offset  will  be  required  for  dead  center  to  turn  the 
taper  when  whole  length  of  pin  is  10  j%  inches? 

23.  The  reamer  to  finish  seat  for  the  pin  in  crank  of 
problem  22,  was  8  in.  long.    What  was  the  offset  of  the  dead 


108  SHOP  MATHEMATICS 

center  to  turn  the  taper  on  a  reamer  when  the  small  dia.  of 
taper  was  4^\  in.  dia.  and  taper  was  6^  in.  long? 

24.  An  axle  turning  lathe  was  equipped  with  high  speed 
tools  for  turning  at  100  F.  P.  M.    What  R.  P.  M.  should  the 
spindle  revolve  for  turning  an  axle  3£  in.  diameter? 

25.  A  lathe  is  fitted  to  turn  heavy  cast  iron  pulleys  at 
60  F.  P.  M.    What  approximate  R.  P.  M.  will  spindle  make 
for  a  30  in.  dia.  pulley? 

METAL  PLANER 

A  sketch  of  the  driving  mechanism  for  a  Powell  planer  is 
shown  in  Fig.  50.    Two  belts  connect  the  countershaft  with 


Main  ttne 


l6Dia-IOaT 


the  driving  shaft,  the  cross  belt  driving  the  platen  forward 
for  the  cut  and  the  straight  belt  causing  the  return.  The 
motion  is  transmitted  from  the  driving  shaft  A  to  the  platen 


SHOP  MATHEMATICS  109 

by  means  of  a  system  of  gearing  as  shown  in  the  cut,  the  last 
gear  meshing  with  a  rack  on  the  under  side  of  the  platen. 

The  cutting  speed  for  planer  tools  is  about  the  same  as 
for  lathe  tools  but  heavier  cuts  can  be  taken  on  account  of 
the  rigid  support  of  work  on  the  platen.  One  great  dis- 
advantage with  doing  work  on  a  planer  is  that  the  return 
stroke  is  time  lost  for  the  tool  in  the  cut,  although  belted 
to  return  about  three  times  as  fast  as  the  cutting  stroke. 
The  calculations  for  the  planer  are  to  determine  sizes  of 
pulleys  to  give  required  travel  of  work  in  F.  P.  M.  The 
formula  is  essentially  the  one  for  drivers  and  followers. 

Example.  What  is  the  travel  of  platen  B,  Fig.  50,  for  one 
revolution  of  shaft  A  ? 

Solution.     By  formula: 

DriversXR  of  1st  driver 

W  —  r  _  ,  _  !L  _ 

Followers 

n  =  revolution  of  last  follower, 
1X15X13      13 


The  travel     =  R.  P.  M.  X  *  X  diameter  of  last  follower 
or,        .024X3.1416X18.75  =  1.4137  inches. 

PROBLEMS 

1.  Find  the  F.  P.  M.  for  the  cut  on  a  planer  when  the 
16  in.  dia.  pulley  on  main  line  runs  170  R.  P.  M.  and  is  belted 
to  a  12  in.  follower  on  countershaft  on  which  is  another  12  in. 
dia.  pulley  belted  to  an  18  in.  dia.  pulley  on  shaft  A,  the 
gearing  from  A  to  platen  being  the  same  as  given  in 
example  illustrated  above. 


110  SHOP  MATHEMATICS 

2.  What  is  the  return  stroke  in  F.  P.  M.  when  the  pulley 
on  countershaft  is  changed  from  12  in.  to  20  in.  dia.  belted 
to  an  1 H  in.  pulley  on  shaft  A  in  place  of  the  18  in.  dia.  pulley 
of  problem  1,  with  the  gearing  from  A  to  platen  remaining 
the  same. 

3.  When  the  drive  pulley  on  main  line  is  changed  from 
16  in.  dia.  to  20  in.  dia.,  otherwise  the  same  as  problem  1, 
what  is  the  F.  P.  M.  of  the  cut? 

4.  If  the  belt  from  driver  on  main  line  to  countershaft 
has  a  slip  of  2%  and  the  one  from  countershaft  to  shaft  A 
has  a  creep  of  4%,  what  is  the  speed  of  problem  1  in  F.  P.  M.f 

5.  What  is  the  time  required  to  take  150  strokes  on  a 
plate  24  in.  long  when  the  travel  in  F.  P.  M.  is  the  same  as 
in  problems  1  and  2,  and  3  in.  travel  is  allowed  at  the  begin- 
ning and  end  of  stroke  for  shipping  the  belt? 

Figure  51  shows  an  outline  of  the  gearing 
for  feeding  the  tool  carrier  across  the  plane 
of  the  work.     The  tool  carrier  is  moved  by 
a  £   inch  lead  square  thread  screw  S,  to 
which  is  fastened  a  19  tooth  gear  C  in  mesh 
J^ra.  5J.      w^h  a  gear  B,  having  78  teeth.    The  ratchet 
wheel    A    with  80   teeth   is   connected  by 
a  rod  to  the  adjustable  cam  that  imparts  a  rotary  motion 
to  it  at  each  stroke  of  the  planer;  fast  to  gear  B  is  the  ratchet 
P  which  holds  the  gear  B  from  moving  back  with  the  rod 
when  it  returns  back  at  end  of  stroke  for  the  next  move- 
ment for  the  feed.     The  feed  motion  is  calculated  by  the  for- 
mula used  for  gear  trains,  page  12.    For  one  tooth  movement 

,          .        78X.250in.Xl 

of  ratchet  wheel  A   the  feed  of  carrier  = 

jy  x  oU 

.01283  inch. 


SHOP  MATHEMATICS  111 

7 '8X. 250X2 
lor     two     teeth     movement,    the    leed  = — — — — —  = 

iy  x  oO 

.02566  inch. 

6.  When  a  planer  mechanism  is  arranged  as  in  problems 
1  and  2  and  the  feed  is  for  a  two  teeth  movement  of  ratchet 
as  shown  in  Fig.  51,  what  is  the  time  required  to  plane  a  plate 
12  in.  square  when  the  time  allowance  for  shipping  is  the 
same  as  in  problem  5? 

7.  What  is  the  time  required  to  finish  a  surface  24  in. 
long  by  18  in.  wide  when  the  gearing  is  the  same  as  in  prob- 
lems 1  and  2  with  8  tooth  movement  of  A  ? 

8.  What  gears  should  be  used  at  B  and  C  to  give  a  .01  in. 
feed  to  the  tool  for  1  tooth  movement  of  A  ? 

9.  What  changes  will  be  required  in  the  sizes  of  two 
driving  pulleys  on  main  line  and  countershaft  if  pulleys  are 
in  the  same  ratio  as  in  problem  1,  to  give  40  F.  P.  M.  for 
planing  cast  iron,  when  the  other  mechanism  remains  the 
same? 

UNIVERSAL  MILLING  MACHINE 

The  universal  milling  machine  is  used  with  various  attach- 
ments for  taking  special  cuts  on  work,  and  more  especially 
for  cutting  the  teeth  of  mills,  cutters,  gears,  reamers,  drills, 
etc. 

The  most  important  attachment  is  the  index  or  spiral 
head  which  is  used  with  a  dead  center  in  tail  stock,  both 
fastened  to  a  bed  or  platen  that  can  be  moved  back  or 
forward  in  a  horizontal  plane  under  a  revolving  cutter  carried 
by  the  main  driving  spindle  of  the  machine.  The  feeding 
mechanism  for  the  platen  is  shown  in  Fig.  52.  A  feed  belt 
connects  the  main  driving  spindle  with  an  intermediate  shaft 
on  which  is  a  stepped  cone  belted  to  a  second  cone  pulley  in 


112 


SHOP  MATHEMATICS 


reverse  order  of  steps  allowing  for  changes  in  speeds  of  the 
feed  gearing;    from  the  second  cone,  the  feed  is  connected 

63T- 


Fig.  52. 

to  the  bed  by  a  system  of  shafts  and  bevel  gears.  This  shows 
a  simple  system  of  connecting  the  feeding  of  the  bed  from 
the  driving  spindle  but  it  is  done  in  a  variety  of  ways.  The 
calculations  for  the  speeds  and  feeds  of  the  milling  machine 
are  almost  identical  with  those  for  the  lathe,  therefore  the 
mathematical  work  of  the  miller  will  be  confined  to  the  index 
head  attachment,  in  connection  with  such  parts  of  the  feed- 
ing mechanism  as  are  related  to  it. 

THE  INDEX  HEAD 

The  index  head  is  used  to  divide  the  periphery  of  a  piece 
of  work  into  &ny  number  of  equal  parts  and  to  hold  the  work 
at  these  positions  to  allow  cuts  to  be  taken  at  equal  intervals, 


SHOP  MATHEMATICS 


113 


as  in  gear  cutting,  etc.    The  mechanism  for  indexing  is  shown 
in  Fig.  53,  with  the  body  and  supporting  parts  removed. 


The  work  which  is  to  be  divided  into  equal  divisions  is  sup- 
ported and  revolved  on  centers  in  much  the  same  manner  as 
work  is  turned  in  a  lathe. 

A  latch  pin  5  is  held  by  the  tension  of  a  spring  in  one  of 
the  holes  of  the  index  plate  6;  when  the  pin  is  pulled  out  of 
the  plate,  crank  4,  can  be  turned,  transmitting  motion 
through  worm  shaft  3,  and  its  worm,  3^  to  worm  wheel  2  fast  to 
spindle  1,  in  which  is  the  live  center  that  supports  the  work 
to  be  cut.  The  worm  wheel  2  has  40  teeth  driven  by  a  single 
thread  worm  3lt  so  that  one  revolution  of  worm  3l}  which  is  also 
the  same  as  one  revolution  of  index  crank  4,  will  move  the 
spindle  1  carrying  the  work  through  ^  of  a  revolution; 
then  forty  revolutions  of  crank  4  will  give  the  work  one 


114  SHOP  MATHEMATICS 

complete  revolution.  If  then  40  teeth,  or  divisions,  are  required 
on  the  work,  one  revolution  of  the  crank  will  be  made  be- 
tween each  cut  as  the  work  is  moved  under  the  cutter  on 
the  main  driving  spindle.  If  20  teeth,  or  divisions,  are 
required,  the  crank  will  be  turned  twice  around  for  each  cut 
since  20  is  one-half  of  40,  10  divisions  will  require  four  turns 
of  the  crank;  8  divisions  five  turns;  5  divisions  eight  turns, 
etc.  From  this  is  obtained  the  formula  for  any  required 
number  of  divisions. 

Let      N=  number  of  divisions  required. 

R  —  number  of  turns  of  the  crank  for  each  cut. 

Then  R  =  ^. 

Example.     Find  the  indexing  required  for  cutting  a  gear 
having  60  teeth. 

Solution.      By  formula:      ^=T7  =  ™=§;    therefore    any 

1\      bO 

plate  having  the  number  of  holes  in  a  row  divisible  by  3  may 
be  used.  In  this  case  take  the  39  hole  index  and  for  each  one 
of  the  60  teeth  cut  move  the  index  pin  5  around  26  holes, 
since  26  =  f  of  39. 

The  length  of  the  crank  4  is  adjustable  to  any  size  of  the 
plate  6,  but  to  avoid  using  a  plate  of  a  large  diameter,  three 
detachable  plates  are  supplied  with  each  machine  so  that 
they  can  be  changed  easily  to  allow  any  simple  indexing  to 
be  made.  The  following  list  gives  the  usual  number  of 
circles  of  holes  on  the  three  plates. 
Plate  Number  of  holes  in  rows: 

1  =  15—16—17—18—19—20 

2  =  2 1—23— 27— 29— 3 1—33 

3  =  37—39—41—43—47—49 


SHOP  MATHEMATICS  115 

A  sector  7  is  adjustable  so  that  its -two  arms  can  be  moved 
to  take  the  space  between  any  two  holes  of  a  series,  thus 
avoiding  the  necessity  of  counting  the  number  of  holes  for 
each  division  as  it  is  spaced  on  the  work. 

PROBLEMS 

1.  Find  the  indexing  for  3  divisions. 

2.  If  a  mill  is  to  have  32  teeth,  what  is  the  indexing 
required? 

3.  If  a  tap  is  to  have  4  flutes  cut  in  the  universal  miller, 
what  indexing  can  be  used? 

4.  A  gear  is  to  have  24  teeth.     What  is  the  required 
indexing? 

5.  A  mill  is  to  have  35  teeth  cut  on  miller.     What  is 
the  indexing  required? 

6.  Find  the  indexing  required  for  45  divisions. 

7.  Find  the  indexing  required  for  72  divisions. 

8.  Find  the  indexing  required  for  68  divisions. 

9.  If  82  teeth  were  required  to  be  cut  on  a  gear  in  the 
universal  miller,  what  indexing  would  be  made? 

10.  A  gear  is  to  have  108  teeth.     Find  the  indexing  re- 
quired. 

11.  An  index  plate  with  116  holes  is  to  be  drilled  on  the 
milling  machine  with  horizontal  drilling  attachment.    What 
is  the  indexing  required? 

12.  A  ratchet  wheel  is  to  have  148  teeth  to  be  cut  on  its 
periphery.    Find  the  required  indexing. 

13.  A  circular  plate  is  to  be  marked  on  the  circumference 
with  164  divisions.    What  indexing  will  allow  the  work  to  be 
done  on  the  universal  miller? 

14.  A  knurl  is  to  have  teeth  ^  in.  circular  pitch  at  bot- 
tom of  cut  on  the  circumference;    if  the  blank  is  3.3706  in. 


116  SHOP  MATHEMATICS 

dia.  and  the  teeth  are  ^  m-   deep,  what  indexing  will  be 
necessary  for  cutting  the  teeth  in  the  universal  miller? 

15.  A  thin  disc  is  to  have  165  notches  cut  on  its  edge  at 
equal  intervals  around  its  circumference.     If  the  work  is 
done  in  universal  miller,  what  indexing  will  be  necessary? 

16.  A  worm  wheel  blank  3.6330  in.  dia.  is  to  mesh  with  a 
TV  in.  lead  V  thread  screw.     What  indexing  will  allow  the 
wheel  to  be  cut  on  miller? 

17.  An  indexing  disc  blank  4.1253  in.  dia.  for  the  planer 
centers  has  V  cuts  in  the  circumference  .06  in.  wide  from 
point  to  point.    Find  the  indexing  to  cut  the  teeth  in  a  uni- 
versal miller. 

18.  A  disc  31.2  in.  in  periphery  has  lines  TV   in.   apart 
on  the  circumference.     What  indexing  will  allow  the  mark- 
ing to  be  done  in  miller? 

19.  A  disc  1  in.  in  periphery  has  lines  at  •£%  in.  intervals 
around  the  circumference.     Find  the  indexing  necessary  to 
do  the  work  in  the  universal  milling  machine. 

COMPOUND  INDEXING 

When  simple  indexing  will  not  give  the  required  spacing 
compound  indexing  may  be  used  as  follows: 

Example.     Find  the  indexing  required  to  cut  69  teeth  in 
a  gear. 

Solution.  (1)  69  =  23X3 

(2)  33—23=10  = 
(3) 

(4)  .40  =  2X2X2X5 

.    (5)  33=11X3 

(6)  23=23X1 


SHOP  MATHEMATICS  117 

(1)  Set  down  the  number  of  divisions  required  and  re- 
solve into  factors. 

(2)  Choose  an  index  plate  as  33  and  23  holes  and  set  their 
difference  under  the  required  number  and  factor  as  in  (1). 

(3)  Draw  a  line  under  the  two  sets. 

(4)  Set  down  the  number  representing  teeth  in  worm 
wheel  and  factor  as  in  (1). 

(5)  Set  down  the  number  of  the  larger  index  and  factor 
as  in  (1). 

(6)  Set  down  the  number  of  the  smaller  index  and  factor 
as  in  (1). 

(7)  Cancel  factors  above  the  line  with  the  factors  below 
and  if  all  the  factors  above  the  line  cancel,  then  the  right  index 
plate  has  been  chosen.    If  the  factors  above  the  line  do  not 
all  cancel,  other  numbers  representing  sets  of  holes  in  the 
index  plates  must  be  chosen  until  a  number  is  found  that 
will  permit  cancellation  of  all  factors  above  the  line.     The 
product  of  the  factors  below  the  line  remaining  uncancelled, 
2X2X11  =  44,  will  be  the  number  required  to  give  the  right 
indexing  fraction;  that  is,  turning  the  33  hole  index  44  holes 
in  one  direction  and  the  23  hole  index  44  holes  in  the  other 
direction  will  make  the  movement  of  the  index 

IHH£ofatum- 

If  any  whole  number  is  subtracted  from  both  of  these 
fractions  the  resulting  fraction  is  unchanged.  In  this  case 
subtracting  one  turn  from  each  fraction  the  indexing  will  be 
|^  of  a  turn  in  one  direction  and  ^  of  a  turn  in  the 


118  SHOP  MATHEMATICS 

opposite  direction.  It  does  not  matter  in  which  direction 
the  indexing  is  done,  as  it  will  simply  turn  the  work  in  either 
direction  throughout  the  cut. 

It  will  be  seen  that  pin  16  is  not  adjustable  and  is  in  line 
only  with  the  outside  row  of  holes  in  index  plates,  therefore 
the  outside  row  of  holes  of  one  of  the  three  plates  must  be 
chosen  for  the  larger  number  for  the  compounding. 

PROBLEMS 

20.  Find  compound  indexing  required  for  77  divisions. 

21.  What  indexing  will  be  required  to  cut  a  gear  of  91 
teeth  in  the  universal  milling  machine? 

22.  A  ratchet  wheel  is  required  to  have  96  teeth  spaced 
at  equal  intervals  on  its  circumference.    What  indexing  will 
be  required  to  do  the  work  in  the  universal  milling  machine? 

Note.  In  problem  22  it  is  possible  to  use  simple  indexing  for 
48  divisions  and  cut  around  once,  then  turn  the  work  one-half  the 
distance  between  two  cuts  and  cut  around  again. 

23.  A  gear  with  99  teeth  is  to  be  cut  in  milling  machine. 
Find  the  required  indexing. 

24.  What    compound   indexing   can   be   made   for    147 
divisions? 

25.  A  gear  is  to  be  cut  in  the  milling  machine  to  have 
154  teeth.    What  indexing  will  be  required  to  do  the  work? 

26.  What  indexing  in  the  milling  machine  will  be  re- 
quired to  do  the  cutting  of  174  equal  divisions? 

27.  Find  the  indexing  required  for  182  equal  spaces  on 
the  circumference  of  a  disc. 

28.  What  indexing  will  cut  186  teeth  in  a  gear? 


SHOP  MATHEMATICS  119 

DIFFERENTIAL  INDEXING 

The  method  of  compound  indexing  has  been  to  a  large 
extent  displaced  by  the  differential  system  of  indexing  that 
is  now  a  special  feature  of  all  the  Brown  and  Sharpe  Manu- 
facturing Company's  universal  milling  machines.  With  the 
differential  indexing  any  number  of  divisions  from  1  to  382 
may  be  obtained  in  the  same  manner  as  for  plain  indexing 
except  that  properly  selected  change  gears  are  used. 

In  plain  indexing  the  index  plate  is  held  firmly  in  position 
by  means  of  pin  16,  Fig.  53,  so  that  ^  of  a  turn  of  the  index 
crank  moves  the  work  on  centers  £  of  ?V  =  T^  of  a  turn. 
In  differential  indexing  the  index  plate  is  connected  with 
the  work  spindle  through  a  train  of  gears. 

Every  movement  of  the  index  crank  causes  a  movement 
of  the  index  plate,  and  this  motion  may  be  either  forward  or 
backward  as  desired. 

The  effects  of  these  motions  may  be  seen  from  the  follow- 
ing illustrations:  Fig.  A  represents  the  method  of  plain  in- 
dexing, with  index  plate  fixed. 


\  of  a  turn  of  the  index  crank  from  M  to  0  between  the 
arms  of  the  sector  produces  ^  of  -£G  =  T;VV  °f  a  turn  of  the 
work  on  work  spindle  1,  Fig.  53.  Fig.  B  represents  the 
result  when  index  plate  is  geared  to  move  forward.  While 
the  index  crank  is  being  moved  through  £  of  a  turn,  from 


120  SHOP  MATHEMATICS 

N  to  0  of  sector,  the  index  plate  is  moving  forward  the  dis- 
tance N  M  so  that  the  work  is  actually  turned  through  more 
than  ^  of  4*0  of  a  turn. 

Fig.  C  represents  the  result  when  the  index  plate  is  geared 
to  move  backward.  While  the  index  crank  is  being  moved 
through  3  of  a  turn  as  indicated  by  the  distance  N  0  be- 
tween the  arms  of  the  sector,  the  index  plate  is  being  moved 
in  the  opposite  direction,  so  that  the  movement  of  the  work 
on  the  work  spindle  centers  is  thus  less  than  y^  of  a  turn. 

The  amount  of  rotation  of  the  index  plate  may  be  regulated 
by  the  difference  in  the  velocity  ratios  of  the  change  gears. 

The  gears  furnished  with  the  Brown  and  Sharpe  Mfg.  Co. 
milling  machine  for  differential  indexing  and  spiral  milling 
cuts  are  as  follows:  3  gears  of  24  teeth  and  one  each  with 
28,  32,  40,  44,  48,  56,  64,  72,  86,  and  100  teeth  respectively. 

Example.     Find  the  indexing  required  for  53  divisions. 

Solution.  By  simple  indexing  the  index  crank  would  be 
revolved  through  |§  of  a  turn  for  each  division,  but  as 
there  is  no  index  plate  having  53  holes  this  spacing  is  im- 
possible, therefore  another  fraction  is  selected  whose  value 
is  near  £§,  say  M  =  |,  then  the  21  or  49  hole  index  plate 
can  be  used,  as  15  holes  in  the  21  row  or  35  holes  in  the 
49  row  gives  the  required  fraction.  Indexing  in  this  way 
for  53  divisions  gives  5<?Xf  =  ^fj&  =  <?7f  complete  turns  of 
the  index  crank  or  2\  turns  less  than  the  40  required  for 
one  complete  turn  of  the  work.  By  using  gears  in  the  ratio 
of  2\  to  1,  however,  the  index  plate  will  make  2\  revolu- 
tions which,  with  the  <?7f  turns  of  the  crank  will  make  the 
40  turns  required.  Hence  the  gears  will  be  in  the  ratio  of 
40X72 


=  _ 
7     7X1        56X24' 


SHOP  MATHEMATICS 


121 


Fig.  54 

Then  place  gear  of  56  teeth  on  worm  as  C,  Fig.  54 
gear  of  40  teeth  first  on  stud 
gear  of  24  teeth  second  on  stud 
gear  of  72  teeth  on  spindle  as  E,  Fig.  54 


Fig.  55 


122 


SHOP  MATHEMATICS 


As  the  motion  of  the  index  plate  is  to  be  in  the  same  direc- 
tion as  the  motion  of  the  crank,  no  idler  gear  will  be  required, 
and  the  setting  to  use  the  compound  stud  gears  is  shown  in 
Fig.  55. 

Example.     Find  the  indexing  for  352  divisions. 

Solution.  By  simple  indexing  the  spacing  number  =  £/?. 
As  this  fraction  cannot  be  obtained  directly  from  the  index 
plate,  select  3\°o  =  ^  .  This  may  be  found  by  using  3  holes 
in  the  27  hole  row  or  2  holes  in  the  18  hole  row;  then  £X 
352  =  39$,  which  is  the  number  of  turns  of  the  index 
crank,  or  f  less  than  the  required  40  for  one  complete 
turn  of  the  work.  The  ratio  of  gears  used  will  therefore 


h        - 

6 


9  9X1 
Example. 
Solution. 


72X24' 

Find  the  indexing  required  for  51  divisions. 
The    spacing    number  =  fy.     As   this   fraction 

cannot  be  taken 
directly  from   the 
index    plate,    select 
M  =  U;   then  the 
index    may 
by     taking 


17 
be 
14 


Fig.  56 


hole 

used 

holes 

which   is  the  number 

of  turns  of  the  index 

crank  or  2  more  than 

the    required    40    for 

one  turn  of  the  work. 

The    ratio     of    gears 

must  then  be  as  2  to 


7      .L  A  -I 

Then  gear  on  spindle  will  have  48  teeth,  gear  on  worm 
will  have  24  teeth. 


SHOP  MATHEMATICS  123 

As  the  motion  of  the  index  plate  must  be  in  the  direction 
opposite  to  the  movement  of  the  index  crank,  two  idler  gears 
must  be  used,  as  shown  in  Fig.  56. 

PROBLEMS 

29.  What  is  the  indexing  required  for  227  divisions? 

30.  What  is  the  indexing  required  for  57  divisions? 

31.  Find  the  indexing  by  the  differential  method  for 
149  divisions. 

32.  Find  the  indexing  required  for  159  divisions. 

33.  What  is  the  differential  indexing  for  161  divisions? 

34.  What  is  the  differential  indexing  for  162  spaces? 

35.  When  the  circumference  of  a  disc  is  to  be  marked  into 
359  equal  spaces,  what  is  the  indexing  required  on  the  index 
head  of  milling  machine? 

THE  SPIRAL  HEAD 

The  index  head  is  used  in  connection  with  the  feeding  of 
the  platen  carrying  the  work,  when  cutting  spiral  or  helical 
flutes  in  reamers  and  drills  or  the  teeth  of  spiral  mills,  gears, 
etc.,  as  well  as  in  indexing  for  the  equal  spaces.  The  object 
is  to  impart  a  rotary  motion  to  the  work  as  it  is  fed  along 
under  the  cutter,  which  is  done  by  the  action  of  a  train  of 
gears  in  mesh  with  the  feed  screw  15,  and  the  worm  3l ,  Fig. 
53.  This  action  is  very  similar  to  that  of  the  gearing  used  in 
thread  cutting  in  the  lathe;  in  fact,  the  flute  of  a  drill  is  a 
thread  of  quite  long  lead,  and  is  spoken  of  as  of  so  many 
inches  to  one  turn,  instead  of  so  many  threads  per  inch  as 
in  the  case  of  a  screw  thread;  thus  a  drill  is  said  to  have  a 
flute  one  turn  in  six  inches,  or  it  might  be  said  to  be  a  flute 
of  six  inch  lead. 


124  SHOP  MATHEMATICS 

It  is  usually  the  custom  to  call  the  cutting  of  the  helical 
flutes  of  straight  reamers  and  drills  and  teeth  in  mills,  gears, 
etc.,  spiral  cuts,  whereas  a  spiral  is  the  winding  cut  on  the 
cone,  but  the  action  of  feeding  the  work  is  the  same  in  the 
case  of  a  straight  or  taper  reamer,  hence  the  custom.  The 
principle  is  as  follows:  Feed  screw  15,  is  a  ^  inch  lead 
double  square  thread  screw.  Removing  pin  16  from  plate  6 
allows  it  to  revolve  through  the  action  of  the  gears  10  and 
13,  8,  9,  etc.  Now,  if  gears  10  and  13  are  the  same  size  and 
gears  11  and  12  are  the  same  size,  or  are  replaced  by  a  single 
idler  gear,  then  forty  turns  of  15  will  give  forty  turns  to  shaft 
3,  which  gives  the  work  carried  by  1  one  complete  turn,  since 
gears  8  and  9  are  miter  gears.  Now  forty  turns  of  15  will 
feed  the  platen  carrying  the  wrork  a  distance  of  Ifi  multiplied 
by  the  lead  of  screw  15,  or  40 X  i  =  10  inches.  This  motion  of 
platen  when  the  ratio  of  gears  is  1  to  1  is  called  the  lead  of 
the  machine  and  is  the  factor  used  to  get  the  ratio  of  gears 
for  any  other  lead  required. 

Let  R  =  the  lead  to  be  cut. 

F  =  product  of  follower  gears  of  the  train. 
D  =  product  of  driver  gears  of  the  train. 

„,  10XF 

Then  R  =  — = — 

or    R:10  =  F:D 
R     F 

and     To=D' 

In  the  arrangement  of  gearing  shown  at  Fig.  53,  gear  13 
is  the  first  driver,  motion  coming  from  the  feed  screw  15, 
the  follower  in  mesh  with  it  is  the  second  gear  that  is  put  on 
the  compound  stud;  the  second  driver  is  the  first  gear  on 
the  compound  stud  and  the  follower  in  mesh  with  it,  gear  10, 


SHOP  MATHEMATICS  125 

the  gear  that  drives  the  worm  3^  through  the  miter  gears 
8  and  9. 

Example.  Find  the  gearing  required  to  cut  a  spiral  flute 
with  a  lead  of  12  inches. 

Solution.    The  gears  10,  11,  12,  and  13  must  be,  by  for- 

12    F 
mula  above,  in  the  ratio  of  12  to  10,  then  j^  =  -f^' 

12    4  X  3 

To  use  the  compound  stud  as  shown,  77;=  ^  —  ^  and     any 

10     o  X  <£ 

4      8     32  3     24 

gears  in  these  ratios  can  be  used,  -^-X-^  =  j^  and  -^-X  —  =. 


72          72y^32     F 

—  then  —  -    —  =  -pr-.     It  makes  no  difference  whether  48  or  40 

4o          4°  X  40     1) 

is  first  driver,  as  long  as  they  are  made  the  drivers  f  or  3  X  4  is 
the  same  as  4  X  3;  also  the  same  principle  applies  to  the  fol- 
lowers, for  2  X  5  is  the  same  as  5  X  2. 

PROBLEMS 

1.  Find  the  gearing  required  for  the  spiral  head  to  cut 
a  flute  of  1  turn  in  13^  inches. 

2.  What  gears  will  be  required  to  cut  the  teeth  in  a 
spiral  mill  when  the  teeth  have  a  lead  of  28  inches? 

3.  What  gearing  will  be  required  to  cut  the  teeth  in  a 
mill  when  the  spiral  has  1  turn  in  22  £  inches? 

4.  Find  the  gears  required  to  cut  26^-  in.  lead. 

5.  What  gears  are  necessary  to  cut  a  48  in.  lead? 

6.  What  is  the  lead  of  spiral  that  will  be  cut  with  gears 
64  on  worm,  48  first  on  stud,  40  second  on  stud,  and  56  on 
screw? 

7.  Find  the  lead  when  gears  72  and  64  are  followers  and 
32  and  40  are  drivers. 


126  SHOP  MATHEMATICS 

8.  What  is  the  lead  of  the  spiral  in  inches  for  1  turn  that 
will  be  cut  with  a  24  tooth  gear  on  worm,  64  first  on  stud, 
48  second  on  stud,  72  on  screw? 

Note.  The  platen  must  be  set  at  the  right  angle  for  the  cut  in 
the  work  otherwise  the  mill  will  cut  on  the  back  of  the  center  of 
revolution.  The  angle  at  which  the  platen  is  turned  is  determined 
by  the  diameter  of  the  work  to  be  cut  and  the  lead  of  spiral,  and  can 
be  found  graphically  or  can  be  calculated  by  trigonometry,  as  the 
circumference  of  the  work  represents  the  side  opposite  the  angle 
and  the  feed  of  bed  is  the  side  adjacent  to  angle. 

Example.  Find  the  angle  through  which  to  swing  the 
platen  of  milling  machine  to  cut  the  flute  in  a  1%  in.  dia. 
drill  with  a  lead  of  12  inches. 


Solution.     Tan  A  =  ~  =  =  .3927  and  this  is  found 

0  l/£ 

in  table,  page  183,  under  tangents  as  21°  30'  for  the  nearest 
angle. 

PROBLEMS 

9.     At  what  angle  should  the  platen  of  a  universal  miller 
be  set  to  cut  £  in  dia.  and  4  in.  lead? 

10.  Find  the  angle  at  which  to  set  the  platen  of  miller 
for  1  £  in.  dia.  and  9  in.  lead. 

11.  At  what  angle  should  the  platen  of  miller  be  set  to 
cut  1-|  in.  dia.  and  8  in.  lead? 

12.  What  is  the  angle  at  which  to  set  the  miller  platen 
for  a  3  in.  dia.  mill  to  cut  a  spiral  tooth  28  in.  lead? 

13.  Find  the  angle  for  3  in.  mill,  one  turn  in  40  inches. 

14.  If  the  spiral  teeth  in  a  3  in.  dia.  roll  are  at  an  angle 
of  30°,  what  would  be  the  length  of  the  roll  for  one  complete 
tooth? 

15.  A  spiral  gear  10  in.  dia.  has  teeth  at  45°  angle.    What 
is  length  for  feed  of  platen  in  miller  for  one  complete  turn 
of  gear  on  centers  of  spiral  head? 


SHOP  MATHEMATICS  127 

16.  A  milling  machine  platen  has  a  feed  of  2  in.  per  min- 
ute and  is  cutting  a  spiral  flute  reamer  $•  in.  dia.  with  8  in. 
lead.    Find  angle  at  which  to  set  the  platen  and  the  time  re- 
quired to  cut  a  flute  of  1^  turns. 

SPEEDS  AND  FEEDS 

The  amount  of  feed  per  revolution  of  a  milling  cutter 
depends  somewhat  on  the  number  of  teeth  in  the  cutter. 
For  backed  off  mills  the  following  table  is  considered  good 
practice: 

Brass,  120  F.  P.  M.,  feed  4f  inches  per  minute,  |  inch  deep. 

Cast  Iron,  60  F.  P.  M.,  feed  4§  inches  per  minute,  £  inch 
deep. 

Wrought  Iron,  48  F.  P.  M.,  feed  2  inches  per  minute,  $  inch 
deep. 

Steel,  36  F.  P.  M.,  feed  1£  inches  per  minute,  |  inch  deep. 

In  the  following  problems  use  the  approximate  formula 

J  C1 

B«~for  R.  P.  M.  unless  otherwise  stated. 
a 

PROBLEMS 

17.  What  speed  is  required  in  R.  P.  M.  for  a  3^  in.  dia. 
milling  cutter  to  be  used  in  cutting  a  steel  forging? 

18.  Find  the  R.  P.  M.  of  a  mill  4^  in.  dia.  for  milling  the 
teeth  of  steel  sprocket  wheel  forgings,  when  cut  is  f  in.  deep, 
and  F.  P.  M.  is  in  inverse  proportion  to  depth  of  cut  as  al- 
lowed in  table  of  feeds  given  above. 

19.  What  R.  P.  M.  will  a  mill  6  in.  dia.  make  at  60  F.  P. 
M.  when  cutting  cast  iron  journal  boxes? 

20.  Find  the  R.  P.  M.  of  a  3  in.  dia.  cutter  for  cast  iron 
surface  plates. 


128  SHOP  MATHEMATICS 

21.  What  R.  P.  M.  should  a  H  in.  T  slot  cutter  revolve 
for  cutting  cast  iron  planer  beds? 

22.  What  speed  in  R.  P.  M.  will  a  £  in.  end  mill  make 
while  slotting  cast  iron  planer  beds  with  £•  in.  depth  of  cut 
if  the  F.  P.  M.  is  in  inverse  proportion  to  depth  of  cut? 

23.  What  speed  in  R.  P.  M.  will  a  5  in.  dia.  mill  make  on 
composition  step  bearings  at  100  F.  P.  M.f 

24.  If  babbitt  metal  bars  can  be  machined  at  130  F.  P.  M. 
find  the  R.  P.  M.  for  a  6  in.  dia.  mill  for  cutting  these  bars. 

25.  An  Ingersoll  milling  cutter  5  in.  dia.  is  finishing  mill- 
ing machine  beds  at  1  in.  feed  per  minute  and  ^  in.  depth 
of  cut.     If  the  mill  makes  60  R.  P.  M.,  exactly  how  much 
faster  in  F.  P.  M.  is  it  cutting  than  is  ordinarily  allowed  for 
cast  iron? 

26.  What  F.  P.  M.  and  R.  P.  M.  should  a  5^  in.  dia. 
milling  cutter  feed  when  taking  a  £  in.  depth  of  cut  in  soft 
brass,  and  the  F.  P.  M.  is  in  inverse  proportion  to  depth  of 
cut? 

27.  How  deep  a  cut  can  a  4  in.  dia.  mill  ordinarily  take 
for  steel  when  running  25  F.  P.  M.  if  the  feed  is  £  in.  per 
minute,  and  F.  P.  M.  is  in  inverse  proportion  to  depth  of  cut? 

28.  What  depth  of  cut  can  be  made  with  a  5  in.  mill  at 
48  R.  P.  M.  in  cast  iron  feeding  4  in.  per  minute  when  feed 
and  depth  of  cut  are  in  inverse  proportion  to  the  F.  P.  M.f 

29.  What  is  the  feed  per  minute  for  a  6  in.  dia.  mill,  £  in. 
depth  of  cut,  for  milling  cast  iron  lathe  cross  slides  at  60 
F.  P.  M.  if  feed  is  in  inverse  proportion  to  the  depth  of  cut? 


SHOP  MATHEMATICS  129 

DRILL  PRESS 

The  feeding  mechanism  of  the  drill  press  is  very  similar 
to  that  of  the  lathe  and  milling  machine,  and  the  cutting 
feeds  and  speeds  for  different  materials  are  about  the  same 
as  for  milling  cutters,  the  speeds  of  drills  being  in  proportion 
to  their  diameters.  The  R.  P.  M.  is  found  by  dividing  the 
constant  for  each  material  by  the  diameter  of  drill. 

Then  R.P.M.  =  -^, 

from  which  the  formulas  are  as  follows: 

on 

for  steel  R.  P.  M.  =  ^, 

for  cast  iron  R.  P.  M.  =  -JT-, 
for  brass         R.  P.  M.  = 


D  =  diameter  of  drill  in  inches. 

Tables  for  speeds  of  drills  have  been  made,  but,  as  in  all 
cutting  operations,  the  kind  and  condition  of  material  and 
tools,  and  the  depth  of  cut  must  always  be  taken  into  account. 
This  can  only  be  determined  by  experience  and  trial.  The 
depth  of  cut  for  small  drills  is  usually  .002  to  .005  inch  per 
revolution,  on  large  drills,  from  .005  to  .015  inch  per  revolu- 
tion, .010  inch  being  a  fair  average  for  machinery  steel. 

PROBLEMS 

1.  Find  the  speed  and  time  required  to  drill  1  in.  cast 
iron  at  .005  in.  feed  per  revolution  with  drill,  f  in.  diameter. 

2.  Find  the  depth  of  cut  for  1  in.  drill  in  a  steel  plate  for 
1  minute  at  .008  in.  feed  per  revolution. 


130  SHOP  MATHEMATICS 

3.  What  is  the  time  required  to  drill  1  in.  deep  in  cast 
iron  with  1£  in.  dia.  drill  at  .012  in.  feed  per  revolution? 

4.  What  is  the  time  required  to  drill  1£  in.  deep  in  brass 
bearings  with  1  1  in.  dia.  drill  at  .009  in.  feed  per  revolution? 

5.  Find  the  time  and  speed  of  a  $  in.  drill  to  cut  a  hole 
|  in.  deep  in  cast  iron  at  .008  in.  feed  per  revolution. 

6.  Find  the  time  and  speed  of  £  in.  drill  to  cut  1  \  in.  deep 
in  cast  iron  at  .005  in.  feed  per  revolution. 

7.  What  is  the  time  required  for  a  1%  in.  dia.  drill  if 
point  is  ^  in.  longer  than  body  size  for  cutting  steel  with  .012 
in.  feed  per  revolution  when  the  work  is  2^  in.  thick? 

8.  What  is  the  time  required  to  drill  a  steel  plate  If  in. 
thick  with  a  2  in.  dia.  drill  when  the  point  is  £  in.  longer  than 
the  body  size  with  a  feed  of  .01  in.  per  revolution? 

9.  Find  the  time  required  to  drill  a  cast  iron  bed  plate 
4  in.  thick  with  a  2  in.  dia.  drill  if  point  of  drill  is  T%  in. 
longer  than  body  size  and  .02  in.  feed  per  revolution. 

HAMMER  BLOW 

The  force  of  a  hammer  blow  is  calculated  from  the  amount 
of  work  done.     In  the  case  of  a  drop  hammer,  the  work  is 
equal  to  the  product  of  the  weight  of  the  hammer  head  and 
the  distance  it  falls.     Neglecting  friction  losses,  the  force  of 
the  blow  may  be  found  by  the  following  formula  : 
Let  F  =  the  striking  force  in  pounds. 
W  =  the  weight  of  the  hammer. 

H=  the  distance  through  which  hammer  moves  in  feet. 
D  =  the  amount  of  compression  of  the  material  or  pene- 
tration expressed  in  feet. 


Then    F- 


SHOP  MATHEMATICS  131 

Example.  A  400  Ib.  drop  hammer  falling  4  ft-  compresses 
a  bar  of  steel  ^  in.  at  each  blow.  What  is  the  force  of  the 
blow? 

Solution.     By  formula: 


12 

PROBLEMS 

1.  A  drop  hammer  weighing  1,000  pounds  falls  3  feet. 
If  the  stock  is  reduced  in  thickness  £  inch  at  the  first  blow, 
what  is  the  force  of  the  blow? 

•  2.  If  the  compression  of  the  material  in  problem  1  is 
lessened  ^4  inch  at  each  succeeding  blow,  what  is  the  force 
of  the  second,  third  and  fourth  blows? 

3.  A  drop  hammer  weighing  575  pounds  falls  from  a 
height  of  30  inches.    If  the  forging  is  compressed  ^  inch, 
what  is  the  force  of  the  blow? 

4.  A  forging  is  made  ^  inch  thinner  at  one  blow  of  a 
425  pound  drop  hammer  falling  a  distance  of  24  inches. 
What  is  the  force  of  the  blow? 

5.  A  10  pound  hammer  falls  a  distance  of  20  inches  and 
compresses  the  material  T^  inch.    What  is  the  force  of  the 
blow? 

6.  Find  the  force  of  the  blow  from  a  weight  of  400  pounds 
falling  a  distance  of  15  feet  that  bends  a  bar  of  steel  \  inch. 

7.  A  pile  driver  has  a  weight  of  500  pounds  which  falls 
20  feet  and  drives  a  pile  12  inches  into  the  ground.    Find  the 
force  of  the  blow. 

8.  What  is  the  force  of  the  blow  in  problem  7  when  the 
pile  is  driven  6  inches? 

9.  When    W=  1,250    pounds,    H  =  3%   feet    and    Z)  =  4 
inches,  what  is  the  striking  force  in  pounds? 


132  SHOP  MATHEMATICS 

HORSE  POWER  OF  MACHINES 

The  horse  power  required  to  run  the  different  machines 
while  taking  the  various  cuts  is  calculated  in  several  ways. 
One  way  is  by  finding  the  speed  and  pull  of  belt,  thus  getting 
directly  the  ft.-lbs.  of  work  done  in  a  specified  time.  This 
calculation  assumes  that  the  cut  taken  is  just  enough  to  use 
all  the  force  of  the  belt  and  not  have  the  pulley  slip.  The 
formula  will  be: 

^WXPXR.   P.  M.X*d 

33,000X12 
when        W  =  width  of  belt  in  inches, 

d  =  diameter  of  pulley  in  inches, 
P  =  pull  of  belt  in  pounds  per  inch  wide,  which  is 
generally  taken  at  about  33  pounds  for 
single  belts  and  50  pounds  for  double  belts. 
Example.     A  3  in.  double  belt  runs  50  R.  P.  M.  on  a  10  in. 
dia.  pulley  of  lathe  and  the  strain  of  belt  is  60  Ibs.  per  in.  in 
width.    What  is  the  H.  P.  of  lathe? 

Solution.     By  formula: 

P  =  3XPPX?PX10X3.1416_31.416  =  „     p 

44 


11          4 

Another,  and  perhaps  a  surer  way,  is  based  on  the  actual 
amount  of  metal  removed  from  the  work  in  chips.  Exhaustive 
experiments  have  been  made  with  the  carbon  steel  cutting 
tools  under  average  conditions,  by  which  a  constant  has  been 
determined  for  the  various  materials. 

Let  W=  weight  of  the  chips  in  pounds  removed  per  hour. 
C  =  constant  which  for  cast  iron  =  .03. 
For  wrought  iron  and  soft  steel  C  =  .032. 
For  crucible  and  tool    steel    C  =  .047. 
Then  formula  isH.P.  =  Cx  W. 


SHOP  MATHEMATICS  133 

Example.     What  is  the  H.  P.  of  a  milling  machine  that 
cuts  50  Ibs.  of  chips  per  hour  from  a  lot  of  castings? 
Solution.     By  formula: 

H.P.  =  CXW  =  50X.03=1.5H.P. 

PROBLEMS 

1.  A  3  in.  single  belt  with  an  effective  pull  of  55  Ibs.  per 
inch  on  9  in.  step  pulley  of  a  milling  machine  runs  50  R.  P.  M. 
What  is  the  H.  P.  of  the  machine? 

2.  A  1£  in.  single  belt  with  an  effective  pull  of  55  Ibs. 
per  inch  runs  142  R.  P.  M.  on  a  10  in.  dia.  pulley  of  a  planer. 
Find  the  H .  P.  of  machine  if  \  the  time  is  on  the  return  stroke. 

3.  A  2  in.  single  belt  on  a  10  in.  dia.  pulley  of  a  drill 
press  runs  60  R.  P.  M.    Find  the  H.  P.  required  to  run  the 
press,  if  the  effective  pull  of  the  belt  is  taken  at  55  Ibs.  per 
inch. 

4.  The  average  F.  P.  M.  for  cutting  work  in  a  planer 
is  10  ft.    If  a  cut  £  in.  deep,  3V  in.  feed,  is  made  on  castings 
12  in.  long,  what  is  the  H.  P.  of  the  machine? 

5.  A  lathe  is  turning  a  2  in.  tool  steel  bar  from  2  in.  to 
If  in.  dia.  in  one  cut  at  25  R.  P.  M.  with  a  ^  in.  feed  per 
revolution.    Find  the  H .  P.  of  the  machine. 

6.  A  milling  cutter  runs  55  R.  P.  M.  with  .02  in.  feed 
per  revolution  on  cast  iron  plates;   the  cut  is  1  in.  deep  and 
3^  in.  wide.    Find  the  H.  P.  required  to  operate  the  machine. 

7.  A  drill  press  with  a  2  in.  dia.  drill  runs  55  R.  P.  M. 
with  feed  .012  in.  per  revolution  in  a  wrought  iron  draw  bar 
plate.    What  H .  P.  will  be  required  to  run  the  press? 

8.  AVhat  H.  P.  will  be  required  to  run  a  lathe  at  30 
R.  P.  M.  turning  a  shaft  to  l^f  in.  dia.  with  a  3V  m- 

and  i  in.  depth  of  cut? 


134  SHOP  MATHEMATICS 

9.  What  H.  P.  will  be  required  to  turn  soft  steel  shaft- 
ing to  6  in.  dia.  with  cut  ^  in.  deep  and  ^  in.  feed  at 
12  R.  P.  M.f 

10.  A  drill  press  cutting  soft  steel  runs  85  R.  P.  M.  with 
a  drill    If  in.    dia.   and   the  feed  .012  in.  per  revolution. 
What  is  the  H.  P.  required  to  run  the  drill  press? 

11.  If  the  cutting  speed  of  a  planer  is  24  F.  P.  M.  what 
H.  P.  will  be  required  to  run  the  planer  on  cast  iron  with  a 
feed  .03  in.  per  stroke,  the  length  of  cut  being  36  in.  and  the 
depth  of  cut  £  inch? 

Note.     In  calculating  the  H.  P.  for  a  planer  the  weight  of  chips 
is  taken  as  though  the  cut  was  continuous. 

12.  Find  H.  P.  to  run  a  milling  machine  in  cast  iron  1  in. 
deep,  1  in.  wide  and  4  in.  feed  per  minute. 

13.  A  milling  machine  takes  a  cut  6  in.  wide,  £  in.  deep 
and  2f  in.  feed  per  minute.     If  the  material  is  soft  steel, 
what  H.  P.  will  be  required  to  operate  the  machine? 

14.  A  drill  press  makes  36  R.  P.  M.  in  a  drop  forging 
with  a  drill  3  in.  dia.  feeding  .015  in.  per  revolution.    What 
H.  P.  will  be  required  to  run  the  drill  press? 

15.  Find  the  H.  P.  of  a  drill  press  running  100  R.  P.  M. 
with  a  2^  in.  single  belt  on  a  12  in.  dia.  pulley. 

16.  Find  the  H.  P.  of  a  lathe  at  70  R.  P.  M.  with  a  4  in. 
single  belt  running  on  a  12  in.  dia.  pulley. 

17.  Find  the  H.  P.  of  a  milling  machine  at  90  R.  P.  M. 
with  a  12  in.  dia.  pulley  having  a  3  in.  double  belt. 

18.  What  is  the  H.  P.  for  operating  a  punch  press  with 
a  4  in.  double  belt  on  a  24  in.  dia.  pulley  running  75  R.  P.  MJ 


SHOP  MATHEMATICS 


135 


DYNAMOMETER 

Another  method  for  finding  the  power  required  to  operate 
machinery  is  by  using  a  dynamometer,  which  is  a  power 
measuring  instrument.  The  accuracy  in  determining  the 
power  is  in  proportion  to  the  kind  and  condition  of  the 
measuring  instrument. 

The  Prony  Brake  is  one  of  the  most  simple  and  familiar 
examples  of  the  dynamometer.  Let  A,  Fig.  57,  represent  a 
pulley  136.04  inches  diameter  that  makes  200  R.  P.  M.,  then 

)0  =  6600  F.  P.  M.    If  a  pull  of  5  pounds  is  made 

at  P  the  work  is  6600X5  =  33,000  ft. -Ibs. 
per  minute  or  1  H.  P.  To  record  the 
amount  of  pull  at  P  is  the  office  of  the 
dynamometer,  as  the  diameter  and  R. 
P.  M.  of  pulley  can  be  readily  obtained. 
Fig.  58  shows  the  general  principle  of 
the  prony  brake  and  the  way  the  appa- 
ratus is  commonly  constructed.  Shoes  a  and  6  can  be 
clamped  to  pulley  with  bolts  c,  c;  when  the  pulley  is  re- 
volved in  the  direction  of  the  arrow,  the  tendency  is  for  the 


A 


entire"]  brake   and    lever   to   rotate  in  the  same    direction, 
which   is   prevented   by   weights  P  in  the  scale  pan   (the 


136  SHOP  MATHEMATICS 

weight  W  is  to  counterbalance  the  weight  of  lever  arm  A 
when  pulley  is  at  rest).  When  the  pulley  revolves  at  its 
normal  R.  P.  M.,  sufficient  weight  P  is  put  in  the  pan  to 
balance  the  lever  between  pins  d,  d,  which  are  placed  to 
prevent  lever  from  revolving.  The  power  absorbed  by  the 
brake  shoes  a,  b,  is  equal  to  the  amount  of  work  which  is 
accomplished  in  ft.-lbs.  per  minute  by  the  revolving  shaft. 
This  work  in  ft.-lbs.  =  PxNx2irXL 

then  H.P.=    "3fl()0  . 

The  brake  must  be  well  lubricated  to  prevent  seizing. 

Example.  An  engine  shaft  makes  150  R.  P.  M.  What  is 
the  H.  P.  developed  when  a  weight  of  10\  Ibs.  is  just 
balanced  at  the  end  of  an  8  ft.  lever  attached  to  a  pair  of 
brake  shoes  as  in  Fig.  58? 

Solution: 

H   P  =  2vLPN=     6.2832X8X10^X150 
33,000  =  33,000 

PROBLEMS 

1.  An  engine  shaft  revolving  at  74  R.  P.  M.  will  support 
a  weight  of  2,000  Ibs.  at  the  end  of  a  10  ft.  lever.    What  is 
the  H.  P.  of  the  engine? 

2.  A  pulley  on  a  motor  shaft  that  revolves  750  R.  P.  M. 
just  balances  a  weight  of  25  Ibs.  on  end  of  a  5'  3"  lever. 
What  H.  P.  is  the  motor  developing? 

3.  A  two  cylinder  gasolene  motor  has  a  fly  wheel  making 
1,000  R.  P.  M.    When  a  5£  ft.  lever  arm  balances  25  Ibs., 
what  is  the  H.  P.  of  the  motor? 


SHOP  MATHEMATICS  137 

4.  What  length  of  arm  will  be  required  to  balance  a  25 
Ib.  weight  on  a  shaft  making  150  R.  P.  M.  doing  2  H.  P.  of 
work,  when  the  brake  shoes  are  clamped  to  the  fly  wheel 
pulley? 

5.  Find  the  weight  required  to  balance  the  brake  shoe 
lever  5'  3"  long  on  a  pulley  at  200  R.  P.  M.  and  transmitting 
10  H.  P. 


138  SHOP  MATHEMATICS 

ENGINE  FLY  WHEELS 

Fly  wheels  are  used  to  regulate  the  motion  in  machinery 
by  storing  energy  during  increasing  velocity  and  giving  it 
out  during  decreasing  velocity.  There  is  no  power  gained  in 
the  use  of  a  fly  wheel,  in  fact,  power  is  used  in  overcoming 
friction  in  bearings  when  the  shaft  is  loaded  with  the 
extra  weight  of  the  wheel.  Fly  wheels  are  usually  made 
of  cast  iron  and  the  safe  velocity  is  one  mile  per  minute 

5280 
or  =  88jeet  per  second. 

The  formula  for  centrifugal  force  in  fly  wheels  is 

.000341. 


gR         3,600g 
W  =  weight  of  rim  in  pounds. 
J^  =  mean  rad.  of  rim  in  feet. 
N  =  R.  P.  M.  of  wheel. 
g  =  32.16. 

V=  velocity  in  feet  per  second  =  2irRN  -r-  GO. 
The  rotating  parts  of  machinery  must  be  well  balanced 
on  their  axes  on  account  of  the  centrifugal  force  which  would 
cause  wear  on  the  bearing  and  vibration  to  the  machinery. 

Example.  What  strain  will  be  put  on  the  bearing  of  a  fly 
wheel  shaft  with  an  unbalanced  weight  of  5  Ibs.  on  the  rim 
of  fly  wheel  which  is  10  ft.  dia.  making  100  R.  P.  M.f 

Solution  : 

10X3.1416X100 


„,,  , 

Thenc./. 


-  =  52.36  feet  per  second. 
60 

5  X  52.36  X  52.36     2741  -5696 


SHOP  MATHEMATICS  139 

The  tensile  strain  (£)  on  the  cross  section  of  rim  is  found 
by  dividing  the  force  by  2v,  then  S=WxRXN2X  .00005427. 
The  maximum  tension  per  square  inch  for  cast  iron,  allow- 
ing factor  of  safety  of  10,  is  1,000  pounds,  corresponding  to 
a  velocity  of  6,085  F.  P.  M.  so  that  one  mile  per  minute  is 
within  the  safe  limit.  The  diameter  of  a  fly  wheel  is  found 
from  formula 

5280 
~  *N 

N  =  R.  P.  M.  of  the  wheel. 
D  =  diameter  of  the  wheel  in  feet. 


PROBLEMS 

1.  When  R  =  7%  ft.,  find  the  strain  in  a  15  ft.  cast-iron 
fly  wheel  rim  that  is  1^  in.  thick  and  16  in.  face,  running 
6,000  F.  P.  M. 

2.  What  extra  strain  will  be  put  on  the  rim  in  problem 
1  if  a  weight  of  10  Ibs.  is  placed  at  some  point  on  the  rim? 

3.  What  strain  will  be  found  in  a  cast-iron  fly  wheel 
4  ft.  dia.  when  the  cross  section  outline  of  rim  is  4  in.  square 
running  at  6,000  F.  P.  MJ 

4.  What  is  the  strain  at  a  speed  of  5,000  F.  P.  M.  of  a 
4  ft.  dia.  cast-iron  fly  wheel  with  a  cross  section  outline  4  in. 
dia.? 

5.  If  R  =  8  ft.,  what  is  the  tensile  strain  in  a  cast-iron 
fly  wheel  16  ft.  dia.  at  74  R.  P.  M.  with  a  rim  which  is  1£  in. 
thick  and  26  in.  wide? 

6.  What  is  the  tensile  strain  in  a  cast-iron  fly  wheel, 
4  ft.  dia.  at  400  R.  P.  M.  with  a  cross  section  outline  of  rim 
4  in.  square? 


140  SHOP  MATHEMATICS 

7.  If  R  =15  ft.,   what  is  the  strain  of  a  cast-iron  fly 
wheel  30  ft.  dia.  with  bolted  rim,  36  in.  width  of  face  and  2£ 
in.  thick,  running  60  R.  P.  M.f 

8.  What  is  the  strain  on  the  rim  of  an  8  ft.  fly  wheel  at 
1 15  R.  P.  M.  with  rim  of  cast  iron  16  in.  wide  and  1  in.  thick? 

9.  If  6,000  F.  P.  M.  is  allowable  for  the  wheel  of  prob- 
lem 7  what  is  the  difference  between  the  actual  and  the  al- 
lowable strain  in  the  rim? 

10.  Find  the  R.  P.  M.  of  a  cast-iron  fly  wheel  8  ft.  dia. 
at  the  safe  velocity. 

11.  Find  the  R.  P.  M.  of  a  3£  ft.  dia.  fly  wheel  at  one  mile 
per  minute  of  rim  speed. 

12.  What  is  the  R.  P.  M.  of  a  14  ft.  fly  wheel  running 
6,000  ft.  per  minute? 

13.  What  is  the  F.  P.  M.  in  an  18  ft.  fly  wheel  rim  running 
at  100  R.  P.  M.? 

14.  How  much  over  the  mile  limit  is  a  22  ft.  fly  wheel 
rim  moving,  when  making  90  R.  P.  MJ 

POWER  OF  THE  STEAM  ENGINE 

There  are  three  kinds  of  horse  power  spoken  of  by  the 
engineer. 

Nominal  horse  power  is  a  term  used  by  makers  of  engines 
as  a  convenient  way  of  stating  the  dimensions  of  an  engine. 
It  really  means  nothing,  is  misleading,  and  should  not  be  used. 

Indicated  horse  power  is  the  actual  measure  of  work  done 
by  the  steam  in  the  cylinder;  this  is  not  based  on  any  as- 
sumption, but  is  actually  calculated  from  the  steam  pressure 
and  the  travel  of  the  moving  parts. 

Effective  horse  power  is  the  amount  of  work  which  an 
engine  is  capable  of  transmitting.  It  is  the  difference  be- 


SHOP  MATHEMATICS  141 

tween  the  indicated  H.  P.  and  the  amount  of  power  required 
to  run  the  engine  without  any  load. 

The  indicated  H.  P.  is  found  by  calculation  and  is  the  one 
generally  meant  when  the  term  H.  P.  of  the  engine  is  used. 

The  H.  P.  of  any  mechanism  depends  on  the  ft.-lbs.  of 
work  done  in  a  specified  time,  therefore  the  power  of  the 
steam  engine  can  be  calculated  in  the  same  manner,  and  the 
usual  formula  taken  is 
PLAN 
33,000 ' 
where  P  =  the  steam  pressure  on  the  piston  called  the 

mean  effective  pressure,  or  M .  E.  P. 
L  =  twice  the  length  of  the  piston  stroke  and  is 
the  travel  of  the  piston  in  feet  for  the  mo- 
tion out  and  back. 

A  =  the  area  of  the  piston  in  square  inches. 
N = the  R.  P.  M.  of  the  crank. 

The  M.  E.  P.  can  be  found  only  by  using  an  indicator, 
but  where  it  is  not  possible  to  use  an  indicator  the  M.  E.  P. 
is  sometimes  taken  at  one-half  the  gauge  pressure  on  the 
boiler. 

Since  the  area  of  the  piston  equals  the  square  of  the  diameter 
X.7854,  then  by  dividing  .7854  by  33,000,  H.  P.  =  D2XPX 
Tx. ,0000238.  There  are  many  formulas  given  for  finding 
the  H.  P.  of  the  steam  engine,  but  all  are  based  on  the  one 
first  given.  A  short  formula  is  sometimes  used  as  follows: 
D'XTXM.E.  P. 


H.  P.= 


42,000 

D  =  diameter  of  piston  in  inches. 
T  =  Lx  N  of  first  formula  given,  which  is  the  dis- 
tance in  feet  of  piston  travel  per  minute. 


142  SHOP  MATHEMATICS 

In  low  pressure  engines,  that  is,  condensing  engines,  it  is 
usually  the  custom  to  add  10  pounds  to  P  for  the  increased 
efficiency  of  the  vacuum. 

Example.     Find  the  horse  power  of  a  steam  engine  when 
the  piston  is  18  in.  dia.,  the  crank  is  15  in.  long  and  makes 
1  15  R.  P.  M.  and  the  M.  E.  P.  is  40  Ibs. 
Solution.     By  formula: 

PLAN  40X5X254.47X115 

°r  H'  P-  ~ 


-  33,000 

The  length  of  crank  is  the  distance  from  the  center  of 
crank  shaft  to  the  center  of  pin.  In  a  complete  revolution, 
the  stroke  being  twice  the  length  of  the  crank,  the  piston 
travel  is  forward  and  back  for  each  stroke  or  5  feet  per  revo- 
lution for  a  15  inch  crank. 

PROBLEMS 

1.  The  dia.  of  a  cylinder  is  16  in.  and  the  stroke  is  24  in.; 
when  the  crank  makes  120  R.  P.  M.  and  the  M.  E.  P.  is  45 
Ibs.  what  is  the  H.  P.  that  will  be  developed  by  a  2-cylinder 
engine? 

2.  Find  the  H  .  P.  of  an  engine  when  the  cylinder  is  13 
in.  dia.,  8  in.  crank,  300  R.  P.  M.  and  M.  E.  P.  67  Ibs. 

3.  Use  the  formula  D2XPX  TX.  0000238  to  find  the 
H.  P.  of  an  engine,  with  18  in.  dia.  of  cylinder,  30  in.  stroke, 
1  15  R.  P.  M.  and  40  Ibs.  M.  E.  P. 

4.  What  is  the  H  .  P.  developed  by  an  engine  with  24  in. 
dia.  cylinder,  60  in.  stroke,  75  R.  P.  M.  of  crank  and  70  Ibs. 
M.  E.  P.? 

5.  Find  the  H.  P.  developed  in  an  engine  with  a  6  in. 
crank,  5  in.  dia.  cylinder  and  400  R.  P.  M. 

Note.  When  the  steam  pressure  is  not  known,  the  H.  P.  will  be 
given  per  Ib.  of  M.  E.  P. 


SHOP  MATHEMATICS  143 

6.  Use  short  formula  for  the  following:   Find  the  H.  P. 
of  a  pair  of  24  in.  dia.,  48  in.  stroke,  horizontal  high  pressure 
cylinders  at  120  R.  P.  M.  of  crank. 

7.  Find  the  H.  P.  of  two  12  in.  dia.  cylinders,  20  in. 
stroke  and  60  R.  P.  M.  of  crank,  when  the  M .  E.  P.  is  to  be 
found  with  an  indicator. 

8.  Find  the  H.  P.  of  an  engine  with  9  in.  dia.  cylinder, 
30  in.  stroke,  90  R.  P.  M.  of  crank,  when  M.  E.  P.  is  to  be 
found  by  an  indicator. 

9.  Find  the  H.  P.  developed  by  a  compound  engine 
when  the  high  pressure  cylinder  is  16  in.  dia.,  M.  E.  P.  70  Ibs., 
and  low  pressure  is  27  in.  dia.  with  M.  E.  P.  12  Ibs.,  stroke 
is  16  in.  and  250  R.  P.  M.  for  each  cylinder. 

Note.  The  H.  P.  of  a  compound  engine  is  found  as  though  each 
cylinder  was  for  a  separate  engine,  by  taking  the  sum  of  the  H.  P. 
of  both  cylinders.  By  the  term  low  pressure  is  meant  that  the  steam 
is  exhausted  from  the  first  cylinder  to  the  second  at  a  lower  pressure. 

10.  Find  the  H.  P.  of  a  compound  engine  when  the  high 
pressure  cylinder  is  27^  in.  dia.,  has  M.  E.  P.  of  36.75  Ibs., 
low  pressure  cylinder  48  in.  dia.,  M.  E.  P.  of  7.25  Ibs.,  length 
of  crank  25  in.  and  75  R.  P.  M. 


POWER  OF  THE  LOCOMOTIVE 

The  term  horse  power  is  seldom  used  in  speaking  of  the 
work  of  the  locomotive  as  the  service  is  entirely  different 
from  the  work  done  by  the  stationary  engine. 

The  power  of  the  locomotive  is  measured  at  the  point  where 
the  wheel  touches  the  rail  and  is  the  weight  on  the  drivers 
and  their  adhesive  power  on  the  rails,  in  other  words  it  is  its 
tractive  force,  or  capacity  to  pull  a  load  by  steam  pressure 


144  SHOP  MATHEMATICS 

through  mechanism  similar  to  the  stationary  engine.  Thus 
a  great  deal  of  the  efficiency  of  the  locomotive  depends  on 
the  speeds,  grades,  curves,  weights  on  drivers,  etc. 

It  has  been  proved  by  actual  test  that  only  43%  of  the 
effective  power  at  the  cylinders  is  available  at  the  draw  bar 
of  a  locomotive. 

Example.  A  locomotive  with  14  in.  dia.  and  24  in.  length 
of  cylinders  and  80  Ibs.  M.  E.  P.  with  6  ft.  dia.  of  drive 
wheels  running  at  one  mile  per  minute  has  what  effective 
power  at  the  draw  bar? 

Solution.     The  H.  P.  at  the  piston  by  formula: 

D      PLAN_80X4X16S.94X280XS_      n  „   p 
~33flOO~  33,000 

43%  of836  H.  P.  =  359. 5  H.  P. 

The  usual  formula  for  the  tractive  power  of  the  locomotive 
is  as  follows: 

Let  T  =  the  tractive  power  of  the  locomotive. 

W  =  the  diameter  of  the  wheel  in  feet. 
S  =  the  length  of  stroke  in  feet. 
D  =  diameter  of  piston  in  inches. 

M.  E.  P.  =  the  pressure  found  with  an  indicator  which 
may  be  taken  as  found  by  experiment, 
to  be  about  40%  of  the  boiler  gauge 
pressure, 

D*XM.  E.  P.XS 

then  T=—       — ==—       —. 

W 

The  tractive  power  is  found  to  be  equal  to  the  load  that 
the  locomotive  can  lift  from  a  pit  by  means  of  a  rope  over  a 
pulley  from  the  circumference  of  the  drive  wheel. 


SHOP  MATHEMATICS  145 

PROBLEMS 

11.  What  is  the  tractive  power  of  a  locomotive  having 
19  in.  dia.  cylinders  and  12  in.  cranks  with  68  in.  drive  wheels 
and  gauge  pressure  160  Ibs.? 

12.  What  is  the  tractive  power  of  a  locomotive  having 
16  in.  dia.  cylinders  with  12  in.  cranks,  72  in.  drive  wheels  at 
180  Ibs.  gauge  pressure  of  which  50%  is  M.  E.  P.?    Also 
what  H.  P.  can  be  applied  at  the  draw  bar  if  only  45%  of 
the  indicated  H.  P.  of  the  piston  is  effective? 

GASOLENE  ENGINES 

The  difficult}'  of  computing  the  horse  power  of  the  gaso- 

PLAN 

lene  engine  with  the  formula  H.  P.  =  —  —  —  is  to  establish 

00,000 

the  M.  E.  P.  and  to  keep  the  piston  speed  at  a  uniform  rate. 

The  A.   L.  A.  M.    (Association  of  Licensed  Automobile 
Manufacturers)  have  adopted  the  following  standard  formula: 

Let         D  =  diameter  of  cylinder  in  inches. 
N=  number  of  cylinders. 

Then  H.P.  =  ^. 
2.5 

Example.     Find  the  H.  P.  of  a  6  cylinder  gasolene  engine 
when  the  cylinders  are  4  in.  diameter. 

Solution.     By  formula: 


The  following  formulas  are  used  by  the  mechanical  en- 
gineers in  some  of  the  leading  automobile  factories  as  giving 
close  results  under  certain  conditions. 


146  SHOP  MATHEMATICS 

SCAN 

12,000  ' 

S  =  stroke  in  inches. 
C=  number  of  cylinders. 
.4.  =  area  of  piston. 
A7  =  number  of  R.  P.  M. 

This  formula  is  based  on  20  cubic  inches  piston  displace- 
ment and  600  R.  P.  M.  =  1  H.  P. 
PLD'N 

H-p'  =    1,000   • 

P  =  M.  E.  P.  taken  as  87  pounds  per  square  inch. 

L  =  length  of  stroke  in  inches. 

D  =  diameter  of  cylinder  in  inches. 

N  =  number  of  cylinders. 

All  of  the  formulas  given  are  based  on  the  effective  work 
that  a  certain  size  of  engine  should  accomplish  and  have  been 
found  correct  in  actual  practice. 

PROBLEMS 

13.  Find  the  M.  E,  P.  of  a  gas  engine  from  formula 

H.  P.  =  —      ^— >  wnen  the  dia.  of  cylinder  is  7|  in.,  length 
00,000 

of  stroke  15f  in.,  making  200  R.  P.  M.,  when  a  brake  test 
of  the  engine  shows  8£  H.  P. 

14.  A  gas  engine  does  25  horse  power  in  a  test  with  a 
6  in.  dia.  cylinder,  12  in.  stroke,  at  600  R.  P.  M.    What  is 
the  M.  E.  P.? 

15.  A  single  cylinder  motorcycle  2|  in.  dia.  and  3  in. 
length  of  stroke  makes  2,500  R.  P.  M.    What  is  the  H.  P. 
whenM.  E.  P.  islOlbs.? 


SHOP  MATHEMATICS  147 

16.  If  the  engine  in  problem  15  has  two  cylinders,  and 
dimensions  the  same,  what  is  the  H.  P.f 

17.  Find  the  H.  P.  by  formula,  H.  P.  =  Sf  ^ '**,  of  a  two 
cylinder  gasolene  engine,  5  in.  stroke,  4  in.  dia.,  1,000  R.  P.  M. 

18.  Find  the  H.  P.  by  formula,  H.  P.  =  P^^N,  of  a  four 

1,000 

cylinder  automobile  engine  making  750  R.  P.  M.,  4£  in.  dia. 
and  stroke  4f  in.  long  with  M.  E.  P.  87  Ibs. 

19.  What  is  the  H.  P.  of  a  six  cylinder  automobile  gaso- 
lene motor  if  D  =  4f  in.  and  L  =  4 \  inches. 

20.  Find  the  H.  P.  of  a  four  cylinder  gasolene  engine 
3f  in.  diameter. 

21.  What  is  the  H.  P.  of  a  six  cylinder  automobile  engine 
with  cylinder  4f  in.  dia.  and  length  of  stroke  4f  inches? 

22.  What  is  the  H.  P.  of  a  single  cylinder  gasolene  motor 
3 1  in.  dia.,  5f  in.  length  of  stroke? 


ENGINE  CYLINDERS 

The  size  of  the  cylinder  can  be  found  by  transposing  the 

formula    for    H.    P.  =  —  -    and    getting   the    diameter 

33,000 

from  the  area. 

PROBLEMS 

23.  Find  the  H.  P.  of  an  engine  when  the  cylinder  is  18 
in.  dia.,  26  in.  stroke,  R.  P.  M.  200,  and  M.  E.  P.  of  48  Ibs. 

24.  What  is  the  H.  P.  of  an  engine  per  Ib.  M.  E.  P.  when 
piston  is  20  in.  dia.,  36  in.  stroke,  175  R.  P.  M.f 


148  SHOP  MATHEMATICS 

25.  A  locomotive's  cylinders  are  16  in.  dia.,  26  in.  stroke, 
drivers  6  ft.  dia.     What  is  the  tractive  power  when  boiler 
pressure  is  200  Ibs.,  M.  E.  P.  taken  at  50%  of  boiler  pressure, 
as  shown  by  the  steam  gauge? 

26.  The  same  as  problem  25  except  boiler  pressure  is 
175  Ibs.  and  M .  E.  P.  taken  at  60%. 

27.  What  size  cylinder  will  be  required  for  an  engine  to 
develop  250  H.  P.  when  the  boiler  pressure  is  100  Ibs.,  and 
M.  E.  P.  taken  at  50%,  24  in.  crank,  at  74  R.  P.  M.f 

28.  What  is  the  H.  P.  of  an  engine  per  Ib.  M.  E.  P.  when 
the  piston  is  18  in.  dia.,  30  in.  stroke,  crank  making  112 
R.  P.  M.f 

29.  Find  the  size  of  a  cylinder  to  develop  75  H.  P.  with 
18  in.  crank  at  115  R.  P.  M.,  boiler  pressure  of  85  Ibs.  and 
M.  E.  P.  taken  at  45%  of  the  boiler  pressure. 

30.  What  dia.  of  cylinder  will  be  required  for  a  150  H.  P. 
engine  with  24  in.  crank,  at  85  R.  P.  M.,  80  Ibs.  boiler  pres- 
sure if  M.  E.  P.  is  taken  at  40%? 

31.  Find  the  size  of  a  cylinder  required  for  a  75  H.  P. 
engine  with  16  in.  crank  at  150  R.  P.  M.  and  M.  E.  P.  37£  Ibs. 

32.  Find  the  length  of  an  engine  crank  when  R.  P.  M. 
is  85,  dia.  of  cylinder  16  in.,  M.  E.  P.  37 £  Ibs.  when  the  en- 
gine is  to  develop  62  H .  P. 

33.  What  is  the  length  and  R.  P.  M.  of  a  crank  for  a 
50  H.  P.  engine,  when  the  piston  is   12  in.  dia.,  and  the 
length  of  crank  is  §  of  the  dia.  of  piston  and  the  M.  E.  P.  is 
30  Ibs.? 

34.  Find  the  H.  P.  of  an  engine  with  a  20  in.  crank  at  140 
R.  P.  M.  with  cylinder  20  in.  dia.  and  M.  E.  P.  49  Ibs. 

35.  Find  the  H.  P.  per  Ib.  M .  E.  P.  with  an  18  in.  crank 
at  130  R.  P.  M.  and  dia.  of  cylinder  16  inches. 


SHOP  MATHEMATICS  149 

INDICATOR  DIAGRAM 

The  M.  E.  P.  on  the  piston  of  an  engine  can  be  found 
accurately  with  an  indi- 
cator.   By  the  mechanism 
of    the     instrument     the 
exact  pressure  on  the 
piston   during   the  entire 
length   of    the    stroke   is 
obtained  in  diagram  on  a  A 
card. 

A  sketch  of  an  indicator 
card  diagram  from  one  end  of  cylinder  is  shown  at  Fig.  59. 

To  find  the  M.  E.  P.  from  the  diagram,  the  line  AB,  which 
is  the  atmospheric  line,  is  drawn  by  the  pencil  of  the  indicator 
when  the  connections  with  the  engine  are  closed  and  each 
side  of  the  piston  is  open  to  the  atmosphere.  Divide  the 
length  of  a  base  line  as  AB  into  any  even  number  of  equal 
parts,  say  10,  setting  off  half  a  part  from  ends  A  and  B  with 
9  parts  between  and  from  these  points  of  division  erect  per- 
pendiculars to  the  base  line  AB  crossing  the  diagram  at  top 
and  bottom;  add  together  the  lengths  of  these  lines  cut  off 
by  the  diagram  and  divide  by  the  number  of  lines.  This 
will  give  the  mean  height,  which  multiplied  by  the  scale 
of  the  spring  used  to  get  the  diagram  will  give  the  M .  E.  P. 

The  area  can  also  be  found  by  Simpsons'  rule,  or  more 
accurately  by  using  a  planimeter. 

STEAM  BOILER 

The  unit  of  H.  P.  for  the  steam  boiler  is  derived  from  the 
number  of  pounds  of  water  evaporated  or  converted  into 
steam  in  one  hour.  The  American  Society  of  Mechanical 


150  SHOP  MATHEMATICS 

Engineers  gives  the  following  rule  which  has  been  adopted 
as  the  standard  unit  for  H.  P.  of  boilers.  1  H.  P.  =  The 
evaporation  per  hour  of  SO  pounds  of  water  from  a  feed  water 
temperature  of  100° F.  into  steam  at  70  pounds  gauge  pressure, 
or  84%  pounds  per  hour  from  and  at  212°. 

From  this  standard  the  capacity  of  a  boiler,  fired  with 
good  anthracite  coal,  to  give  the  above  evaporation  per 
H.  P.  with  maximum  economy,  is  given  by  Kent  as  follows: 

Proportions  of  Grate  and  Heating  Surface 

1.  The  heating  surface  per  H.  P.  =  11%  square  feet. 

2.  The  grate  surface  per  H.  P.  =  %  square  foot. 

3.  The  ratio  of  heating  to  grate  surf  ace =34%  to  1. 

4.  The  water  evaporated  per  square  foot  of  heating  surface 
from  212°F.  =  3  pounds  per  hour. 

5.  The  combustible  burned  per  H.  P.  per  hour =3  pounds. 

6.  The  combustible  burned  per  square  foot  of  grate  sur- 
face per  hour  =  9  pounds. 

7.  The  water  evaporated  at  a  temperature  of  212°F.  per 
pound  of  combustible  =  11%  pounds. 

Example.  What  heating  and  grate  surface  will  be  re- 
quired for  a  200  H.  P.  boiler? 

Solution.  The  heating  surface  by  (1)  is  11%  sq.  ft.  per 
H.  P.,  then  for  200  H.  P.  the  heating  surface  will  be  11% 
X200  =  2,300  sq.ft.  If  %  sq.  ft.  of  grate  surface  is  required 
per  H.  P.  then  for  200  H.  P.  it  will  require  200 X  i  sq.  ft=  66% 
sq.  ft.  of  grate  surface. 

Example.  How  many  Ibs.  of  good  coal  will  be  required 
to  run  a  200  H.  P.  boiler  for  10  hours? 

Solution.  If  3  Ibs.  of  coal  are  required  per  H.  P.  for  1  hr., 
for  200  H.  P.  it  will  require  200X3=600  Ibs.  per  hr.;  and  if 
600  Ibs.  of  coal  are  required  for  1  hr.,  for  lOhrs.  it  will  require 
600X10  =6, 000  Ibs. 


SHOP  MATHEMATICS  151 

TO  FIND  HEATING  SURFACE  FOR  VERTICAL  TUBULAR 

BOILER 

RULE.  Multiply  the  circumference  of  fire  box  in  inches  by 
the  height  above  the  grate;  multiply  the  combined  circum- 
ference of  all  the  tubes  by  the  length  of  one  tube,  in  inches, 
and  to  the  sum  of  these  two  products  add  the  area  of  the 
lower  tube  sheet,  first  subtracting  the  area  of  all  the  tubes 
in  inches ;  divide  the  sum  of  these  products  by  144  to  find  the 
square  feet  of  heating  surface. 

Example.  What  is  the  heating  surface  of  a  vertical 
tubular  boiler  having  twenty-four  3£  in.  dia.  and  8  ft. 
length  of  tubes,  when  the  fire  box  is  16  in.  high  and  boiler 
30  in.  diameter? 

Solution: 

By  rule  cXh  =  30X^X16  =  1508  sq.  in. 
and     3%  X  *  X  24  X  96  =  25,334  sq.  in. 
Area  =  30X30X  .7854  =  707  sq.  in. 
then   1508 +25, 334 +707  =  27, 549  sq.  in. 
and    £7,549  + 144  =  191.3  sq.  ft. 

TO  FIND  HEATING  SURFACE  FOR  HORIZONTAL  TUB- 
ULAR BOILER 

RULE.  Take  all  dimensions  hi  inches.  Multiply  two- 
thirds  of  the  circumference  of  the  shell  by  its  length ;  multiply 
the  combined  circumference  of  all  the  tubes  by  the  lengths 
of  one  tube,  to  the  sum  of  these  two  products  add  two-thirds 
of  the  area  of  both  tube  sheets  and  subtract  twice  the  com- 
bined area  of  all  the  tubes;  divide  this  remainder  by  144  to 
find  the  square  feet  of  heating  surface. 


152  SHOP  MATHEMATICS 

Example.  What  is  the  heating  surface  of  a  horizontal 
tubular  boiler  5  ft.  dia.  and  16  ft.  long  with  fifty  2£  in.  dia. 
tubes? 

Solution: 

,    2XCXI    2X*X60X192 

By  rule =  —        ——      —  =  24127.5  sq.  in. 

o  3 

then       2%X50X-*X192  =  75398.4  sq.  in. 

|  area  of  heads  minus  twice  the  area  of  all  the 
tubes 

or         «X*xa?X-7M*-Htx«>X(t»-X.7«W- 

t> 

3769.92—490.875  =  3279  sq.  in. 

™  24127.5+75,398.4+3279      __  fot 

Then  —  =  713.9  square  feet. 

144 

TO  FIND  THE  HEATING  SURFACE  FOR  ANY  NUMBER 

OF  TUBES 

RULE.  Multiply  the  number  of  tubes  by  the  diameter  of 
one  tube  in  inches,  this  product  by  its  length  hi  feet,  and  this 
product  by  .2618.  The  final  product  will  give  the  square  feet 
of  heating  surface. 

Example.  Find  the  heating  surface  of  fifty-six  3  in.  dia. 
tubes,  18  ft.  long. 

Solution.  By  the  rule,  56X3X18X. 2618  =791. 68  square 
feet. 

TO  FIND  THE  WATER  CAPACITY  OF  A  BOILER 

RULE.  Multiply  two-thirds  of  the  area  of  the  head  in 
square  inches,  by  the  length  of  the  boiler  hi  niches,  and  sub- 
tract the  volume  of  the  tubes  hi  cubic  inches ;  divide  the  re- 
mainder by  23 1 ,  to  find  the  capacity  in  gallons. 


SHOP  MATHEMATICS  153 

Example.  What  is  the  water  capacity  of  a  horizontal 
tubular  boiler  72  in.  dia.  and  18  ft.  long  with  66  tubes  3f  in. 
outside  diameter? 

Solution: 

By  rule  V  =  §  A  X  L  —  volume  of  tubes. 

Then 


586,297—157,457  =  428,840  cubic  inches. 
then  ^j~  =  1856.  4  gallons. 


TO   FIND   THE   STEAM   CAPACITY   OF  A   BOILER 

RULE.  Multiply  one-third  area  of  head  in  square  inches 
by  the  length  of  the  boiler  in  inches,  and  divide  this  product 
by  1728  to  find  capacity  in  cubic  feet. 

Example.  What  is  the  steam  capacity  of  a  horizontal 
tubular  boiler  6  ft.  dia.,  18  ft.  long  with  sixty-six  3£  in.  dia. 
tubes? 

Solution: 

A  y  7 
By  rule      ^+1,7*8, 


then  +i.rn--mji  «,.  ft. 


TO  FIND  THE  PRESSURE  CARRIED  BY  STAY  BOLTS 

RULE.    Find  the  area  of  the  space  be-  > 
tween  any  set  of  adjacent  bolts  as  A,  B,  C, 
D,    Fig.    60,    and    multiply    this    area    in 
square  inches  by  the  pressure  of  steam  gauge. 
This  gives  the  pressure  carried  by  one  bolt;    c,  .1) 

to  find  the  pressure  for  any  number  of  bolts,          ^, .      ^  „ 
multiply  by  that  number.  <-/ 

The  area  of  bolt  must  be  subtracted  for  accurate  results. 


154  SHOP  MATHEMATICS 

TO   FIND   THE   BURSTING   STRENGTH   OF  A  BOILER 


B  =  bursting  strength  of  boiler. 
d=  diameter  of  boiler  in  inches. 
t  =  thickness  of  plate. 
T  =  tensile  strength. 

c  =  coefficient  of  strength  of  the  riveted  joint  and  is 
the  ratio  of  the  strength  of  the  joint  to  the 
solid  plate. 

For  double  riveted  joints  c=  .7. 
For  single  riveted  joints  c=  .5. 

The  safe  working  pressure  for  a  boiler  may  be  taken  as 
follows  : 

2tTc 

"~W 

P  =  safe  working  pressure. 
/=  factor  of  safety  usually  taken  as  6. 
The  horizontal  seams  in  a  boiler  are  double  riveted  unless 
otherwise  stated. 

Example.     Find  the  bursting  strength  and  safe  working 
pressure  of  a  78  in.  dia.  boiler  18  ft.  long  with  f  in.  thickness 
of  plates  tested  at  62,000  Ibs. 
Solution.     By  formula: 

2t  T  c    2  X  f  X  62,000  X  .7    54,250 
B=^T  ~78-  -^-  =  695.5  Ibs. 

and  dividing  by  6  for  the  safe  working  pressure 

695.5      ,f,07, 

—  =  115.9  Ibs. 
b 

The  length  of  a  boiler  is  approximately  3\  times  its  diameter. 


SHOP  MATHEMATICS  155 

PROBLEMS 

1.  Find  the  cost  for  coal  at  $3.50  per  ton  to  run  a  200 
H.  P.  boiler  for  20  days  of  10  hrs.  at  the  average  coal  supply 
for  a  well  equipped  boiler. 

2.  What  is  the  cost  of  water  and  coal  for  running  an 
economically  equipped  power  plant  of  150  H.  P.  for  a  year 
of  300  days  of  10  hrs.  each,  when  coal  is  $3.50  per  ton  and 
water  is  8c  per  100  gallons? 

3.  Find  H.  P.  of  a  vertical  tubular  boiler  48  in.  dia. 
having  thirty  2£  in.  dia.  tubes,  fire  box  20  in.  high  and  tubes 
8  ft.  long. 

4.  What  is  the  H.  P.  of  a  horizontal  tubular  boiler  72  in. 
dia.,  18  ft.  long  with  one  hundred  3^  in.  dia.  tubes? 

5.  What  is  the  heating  surface  of  a  horizontal  tubular 
boiler  20  ft.  long,  5  ft.  6  in  dia.,  with  ninety  3£  in.  dia\  tubes? 

6.  Find  the  heating  surface  of  a    horizontal   tubular 
boiler,  18^  ft.  long,  5  ft.  3  in.  dia.,  having  eighty  3  in.  tubes. 

7.  What  will  be  the  heating  surface  of  a  horizontal 
tubular  boiler,  18^  ft.  long,  5  ft.  2  in.  dia.,  with  seventy-two 
3  in.  tubes? 

8.  Find  the  steam  capacity  for  the  boiler  in  problem  4. 

9.  Find  the  heating  surface   of  a   horizontal  tubular 
boiler  18  ft.  long,  5  ft.  dia.,  with  sixty-six  3  in.  tubes. 

10.  What  is  the  grate  surface  in  sq.  ft.  of  problem  6? 

11.  Find  the  sq.  ft.  of  grate  surface  of  problem  7. 

12.  How  much  more  was  the  cost  of  coal  per  week  of 
56  hrs.  at  4  dollars  per  ton  of  2,000  Ibs.  for  running  a  100 
H.  P.  boiler  with  4,800  Ibs.  of  coal  per  day  of  10  hrs.  than  if 
the  boiler  had  been  operated  on  the  economical  basis? 

13.  If  the  grate  area  had  been  24  sq.  ft.  in  problem  12, 
how  much  should  it  be  increased  to  bring  it  up  to  the  econom- 
ical measurement? 


156  SHOP  MATHEMATICS 

14.  What  is  the  water  capacity  for  a  boiler  18  ft.  long, 
72  in.  dia.,  with  ninety  3£  in.  tubes? 

15.  What  is  the  water  capacity  for  a  boiler  21^  ft.  long, 
6  ft.  dia.,  with  ninety-six  3^  in.  tubes? 

16.  Find  the  steam  capacity  for  a  boiler  16^  ft.  long 
with  one  hundred  3  in.  tubes  when  boiler  is  6  ft.  diameter. 

17.  Find  the  steam  space  for  a  boiler  21  ft.  long  by  78  in. 
dia.  with  one  hundred  3£  in.  tubes. 

18.  What  is  the  bursting  strain  for  a  boiler  48  in.  dia., 
single  riveted,  with  a  \  in.  thickness  of  plate  tested  at  60,000 
Ibs.  per  square  inches? 

19.  Find  the  bursting  pressure  of  a  72  in.  dia.  boiler  for 
^  in.  boiler  plate  tested  at  55,000  Ibs. 

20.  What  should  be  the  thickness  of  plates  for  a  66  in. 
dia.  boiler  to  carry  125  Ibs.  steam  pressure  with  6  for  factor 
of  safety  and  plates  to  be  tested  at  60,000  Ibs.? 

21.  What  is  the  heating  surface  and  bursting  strength 
of  a  horizontal  tubular  boiler  18£  ft.  long,  66  in.  dia.  with 
seventy-two  3  in.  tubes,  when  made  of  £  in.  plates  tested  at 
65,000  Ibs.? 

22.  Find  the  bursting  strength  of  a  boiler  66  in.  dia.  with 
£  in.  plates  tested  at  65,000  Ibs. 

23.  What  is  the  bursting  strength  of  a  boiler  10  ft.  dia. 
with  H  in.  thickness  of  steel  plates  at  80,000  Ibs.  tensile 
strength? 

24.  What  is  the  allowable  pressure  on  a  boiler  7£  ft.  dia. 
made  of  1  in.  steel  plates  tested  at  65,000  Ibs.  tensile  strength? 

25.  WThat  thickness  of  plate  with  a  test  of  65,000  Ibs. 
per  sq.  in.  should  be  used  for  a  6  ft.  dia.  boiler,  if  6  is  the  fac- 
tor of  safety  and  100  Ibs.  per  in.  boiler  pressure  is  to  be  used? 


SHOP  MATHEMATICS  157 

26.  Find  the  thickness  of  the  plate  to  use  for  a  boiler 
66  in.  dia.  to  carry  80  Ibs.  gauge  pressure  when  plates  are 
tested  at  65,000  Ibs. 

Note.  In  all  calculations  for  strength  of  boilers  it  is  assumed 
that  the  riveting  is  of  right  proportions  and  good  workmanship 
throughout.  A  boiler  plate  is  supposed  to  be  tested  at  not  over 
one-third  the  breaking  strain  of  the  plate.  The  tensile  strength  of 
a  stay  bolt  is  not  over  6,000  Ibs.  per  square  in.  and  area  is  taken  at 
the  bottom  dia.  of  thread. 

Example.     What  pressure  is  carried  on  forty  f  in.  stay 
bolts  that  are  spaced  4  in.  on  centers,  when  steam  gauge 
shows  75  Ibs.? 
Solution: 

Area  of  pressure  =  4X4  =  16  sq.  in. 
Pressure  on  one  bolt  =  16X75  =  1,200  Ibs. 
Pressure  on  40  bolts  =1,300X40  =  48, 000  Ibs. 
Area  of  bolt=.6875X .6875X.7854=  .371  sq.  in. 
Area  of  40  bolts  =  .371X40=14.84  sq.  in. 
14.84X75  =  1,113  Ibs. 

Actual  total  pressure  on  40  bolts  =  48, 000— 1,113  =  46,887 
Ibs. 

PROBLEMS 

27.  What  is  the  total  pressure  carried  on  thirty-six  £•  in. 
stay  bolts  spaced  4£  in.  between  centers,  (see  Fig.  60,)  when 
the  steam  gauge  shows  87  Ibs.  pressure? 

28.  Find   the  pressure  on  forty-eight  £  in.  X  10  pi.  stay 
bolts  with  U.  S.  S.  thread  when  boiler  pressure  is  95  Ibs.  and 
area  between  a  set  of  bolts  is  12  square  inches. 

SAFETY  VALVE 

The  lever  safety  valve,  Fig.  61,  is  a  lever  of  the  third  class, 
and  calculations  for  lengths  of  arms  and  weights  required 
for  given  boiler  pressure  are  made  from  the  formulas  of  the 
lever  except  that  weights  of  lever  and  valve  must  be  taken 
into  account. 


158 


SHOP  MATHEMATICS 


When  the  lever  and  valve  connected  to  it  will  just  balance 
over  a  knife  edge,  this  point  is  called  the  center  of  gravity 
of  the  lever;  the  fulcrum  is  at  the  center  of  pivot  on  which 
the  lever  swings. 

Then  let      g  =  the  distance  between  center  of  gravity  and 

fulcrum. 

w  =  weight  of  ball  in  pounds. 
VL  =  weight  of  valve  and  lever  in  pounds. 
A  =  area  of  safety  valve  in  square  inches. 
a  =  distance  between  ball  and  fulcrum  in  inches. 
b  =  distance  between  center  of  valve  and  ful- 
crum in  inches. 

P  =  pressure  per  square  inch  on  steam  gauge. 
The  formula  for  weight  to  balance  the  pressure  is 


and  g,  A,  P,  a,  etc.,  can  be  found  by  transposing  terms  and 
solving  by  algebra;  the  above  formula  can  be  written 

w_APb-VLg 

a 

Example.  What 
distance  from  the 
center  of  fulcrum 
will  a  weight  be 
placed,  if  the  steam 
gauge  shows  95  Ibs., 
weight  is  15  lbsv 
area  of  valve  3 
square  in.,  and 
valve  and  lever 
weigh  18  Ibs.,  cen- 
ter of  valve  2£  in.  from  fulcrum,  0=  12  inches? 


SHOP  MATHEMATICS  159 

Solution.     Substituting  the  values  given  in  the  formula 
3X95X2$—  18X12 

10  =        —r     -• 

3X96X8}— 18X18 

then  a  =  -  —=33.1  in. 

15 


PROBLEMS 

1.  From  the  law  of  levers  show  how  the  formula  for  W 
is  derived. 

2.  What  is  the  weight  for  a  ball  on  a  lever  safety  valve 
5  in.  dia.,  if  the  ball  is  placed  30  in.  from  center  of  fulcrum, 
the  center  of  gravity  is  12  in.  from  fulcrum,  valve  and  lever 
weigh  20  Ibs.,  steam  gauge  registers  80  Ibs.,  and  centre  of 
valve  is  3  in.  from  fulcrum? 

3.  What  is  the  dia.  of  a  cast  iron  ball  for  safety  valve, 
when  the  center  of  gravity  is  12  in.  from  fulcrum,  lever  and 
valve  weigh  14  Ibs.,  distance  from  center  of  ball  to  fulcrum 
is  27  in.,  center  of  2£  in.  dia.  valve  is  3  in.  from  fulcrum, 
and  the  steam  gauge  shows  85  Ibs.  pressure? 

4.  When  the  center  of  a  safety  valve  3.75  in.  dia.  is  2£  in. 
from  fulcrum,  what  distance  from  the  fulcrum  must  a  6^  in. 
iron  ball  be  placed,  if  the  valve  and  lever  weigh  16  Ibs., 
boiler  pressure  is  70  Ibs.,  and  center  of  gravity  is  16  in.  from 
fulcrum? 

5.  What  weight  placed  30  in.  from  the  fulcrum  of  a 
safety  valve  4^  in.  dia.  will  just  balance  80  Ibs.  boiler  pres- 
sure, when  the  valve  and  lever  weigh  12  Ibs.,  the  center  of 
gravity  is  13  in.  from  fulcrum,  and  the  center  of  valve  is  4J 
in.  from  fulcrum? 


160  SHOP  MATHEMATICS 

HYDRAULICS 

Hydraulics  treats  of  liquids  in  motion,  especially  of  the 
action  of  water  in  canals,  pipes,  machinery  for  raising  water 
and  the  use  of  water  as  a  source  of  power. 

Pressure  varies  directly  as  the  depth  from  the  free  surface. 
This  depth  from  the  free  surface  is  called  the  head.  If  the 
weight  of  a  cubic  foot  of  water  is  taken  as  62  £  pounds,  then 
the  weight  of  a  column  of  water  1  foot  high  and  1  square  inch 
in  cross  section  =  62.  5  -r-  144=  -4$4  pound. 

Therefore  the  pressure  per  square  inch  at  any  point  in  a 
body    of   water  =  the    depth    below    the    surface,    or    the 
head,  X.W, 
then  let 

P  =  pressure  per  square  inch. 


Then       P  =  HX.434 
and         H 


To  find  the  head  when  pressure  is  given 

RULE.  Divide  the  pressure  by  .434  or  multiply  the  pres- 
sure by  2.302. 

The  folio  whig  laws  apply  to  liquids  : 

The  pressure  does  not  depend  upon  the  size  or  shape  of  the 
vessel. 

The  pressure  increases  with  the  depth  below  the  free  surface. 

At  any  point  in  a  liquid  the  upward,  downward  and  lateral 
pressures  are  equal. 

The  pressure  which  a  body  of  liquid  exerts  on  the  contain- 
ing vessel  such  as  the  walls  of  a  tank,  is  subject  to  the  follow- 
ing: 


SHOP  MATHEMATICS  161 

RULE.  Pressure  is  equal  to  the  product  of  the  head,  the 
area  of  the  surface  on  which  the  liquid  presses  and  the  weight 
of  a  cubic  foot  of  the  contained  liquid. 

From  this  rule  is  obtained  the  formula: 

P=H  A  W 
Let     W  =  62.5  pounds  =  weight  of  a  cubic  foot  of  water. 

.-1  =  area  in  square  feet  of  surface  in  contact  with 

liquid. 

H  =  the  head,  which  is  the  distance  from  the  free 
surface  of  the  liquid  to  the  center  of  the  sur- 
face in  contact  with  the  liquid. 

Archimedes'  discovery,  that  a  solid  body  immersed  in  a 
liquid  displaces  the  same  volume  as  itself,  furnishes  an  ex- 
cellent method  of  finding  the  volume  of  any  irregular  shaped 
body,  by  immersing  it  in  water  and  measuring  the  volume 
of  the  water  displaced.  The  weight  of  the  body  can  then  be 
found  by  multiplying  the  volume  of  water  displaced  by  the 
weight  of  a  cubic  inch  of  the  substance  of  which  the  body  is 
composed. 

Example.  An  iron  casting  when  immersed  displaces  half  a 
gallon  of  water.  Find  the  weight  of  casting. 

Solution.     A   cu.   in.    of    cast  iron  weighs  .2604  lb.  and 
there  are  115.5  cu.  in.  in  a  half  gal.,  therefore 
115.5  X  .3604  =  30.076  Ibs. 

The  specific  gravity  of  a  substance  is  its  weight  as  com- 
pared with  the  weight  of  an  equal  volume  of  water. 

PROBLEMS 

1.  Find  the  specific  gravity  of  lead  when  a  cu.  in.  weighs 
.4106  Ib. 

2.  Find  the  specific  gravity  of  coal  when  a  cu.  ft.  weighs 
57  Ibs. 


162  SHOP  MATHEMATICS 

3.  Find  the  specific  gravity  of  southern  pine  timber 
when  a  cu.  ft.  weighs  60  Ibs. 

4.  A  cu.  in.  of  a  certain  composition  metal  weighs  .358 
Ib.    What  is  its  specific  gravity? 

5.  What  pressure  will  be  shown  on  the  water  gauge  of 
a  boiler  if  the  supply  tank  has  a  head  of  200  feet? 

6.  What  head  will  show  120  Ibs.  pressure  on  the  water 
gauge  of  a  receiving  tank? 

7.  A  tank  that  is  3  ft.  square  is  filled  to  a  depth  of  3  ft. 
with  water.    What  is  the  pressure  of  the  water  on  the  bottom 
of  the  tank? 

8.  What  is  the  pressure  of  water  on  one  side  of  tank  of 
problem  7? 

9.  A  tank  is  12  ft.  square  on  its  base  and  5  ft.  high. 
What  is  the  pressure  of  the  water  on  one  side,  if  the  tank  is 
filled  to  a  depth  of  5  feet? 

10.  When  the  tank  of  problem  9  is  a  closed  tank  and  a 
pipe  5  in.  dia.  runs  5  ft.  above  top  of  tank,  what  is  the  pres- 
sure on  top  surface  of  tank  when  the  water  is  filled  to  the 
top  of  pipe? 

11.  A  5  in.  dia.  pipe  closed  on  lower  end  is  sunk  in  a  ver- 
tical position  in  a  tank  of  water  to  a  depth  of  30  in.    What 
is  the  upward  pressure  on  the  closed  end  of  the  pipe? 

12.  A  sluice  gate  is  6  ft.  high  and  4  ft.  wide.    When  closed 
the  water  is  up  to  the  top  on  one  side  and  2  ft.  high  on  the 
other.    What  is  the  pressure  on  the  gate? 

13.  When  a  cast  iron  ball  displaces  1£  gal.  of  water, 
what  is  the  weight  of  the  ball? 

14.  If  an  iron  casting  displaces  3  qts.  of  water,  how  much 
does  it  weigh? 


SHOP  MATHEMATICS 


163 


15.  A  steel  bar  was  sunk  in  a  tank  full  of  water  and  it 
was  found  that  15  qts.  of  water  ran  out  to  allow  space  for 
the  bar.    What  was  the  weight  of  the  bar? 

16.  An  irregular  shaped  plate  of  steel  was  immersed  in  a 
tank  of  water  and  found  to  displace  6  gal.     What  was  the 
weight  of  the  plate? 

17.  A  steel  forging  was  found  to  displace  6  qts.  of  water. 
How  much  did  it  weigh? 

18.  How  many  feet  of  lead  pipe  weighing  10  Ibs.  per  ft. 
in  length,  will  displace  10  gal.  of  water? 


HYDRAULIC  MACHINES 

Elevators  and  pumps  are  the  most  common  examples  of 
hydraulic  machines  used  in  shop  practice. 

The  hydraulic  elevator  is  based  on  the  principle  known  as 
Pascal's  law  to  the  effect  that  pressure  exerted  upon  a  liquid 
in  a  containing  vessel  is  transmitted  equally  and  undimin- 
ished  throughout  the  liquid. 

If  pipe  a,  Fig.  62,  has  an  area  of  1  sq.  in.  and  the  pipe  6 
has  an  area  of  100  sq.  in.  and  water 
is  pumped  into  6  through  pipe  a  at  a 
pressure  of  200  Ibs.  per  sq.  in.,  then 
for  each  sq.  in.  of  surface  on  piston 
B  there  will  be  a  pressure  of  200 
Ibs.,  and  on  the  whole  surface  of 
B  there  will  be  a  pressure  of 
100X200  Ibs.  or  20,000  Ibs. 

Example.  A  supply  pipe  for  a  12  in.  plunger  hydraulic 
elevator  piston  is  1  in.  area,  and  the  pressure  into  supply 
pipe  is  pumped  up  to  150  Ibs.  per  sq.  in.  What  is  the  total 
pressure  on  bottom  of  piston? 


Fij 


.62. 


164  SHOP  MATHEMATICS 

Solution,  Since  pressure  on  a  liquid  is  transmitted  un- 
diminished  in  all  directions,  a  pressure  of  n  Ibs.  per  sq.  in. 
on  small  pipe  will  produce  the  same  pressure  per  sq.  in.  on 
large  pipe  at  the  piston,  therefore 

150  :  x  =  1  in.  :  113.10  in.        x  =  16,965  Ibs. 

Pumps  are  of  several  kinds  and  are  operated  by  hand  or 
power.  The  formula  for  lifting  or  forcing  water  either  under 
pressure  or  head  is  as  follows  : 

P  =  H  A  W, 

where    H  is  the  distance  from  the  level  of  the  source  of  sup- 
ply to  the  point  of  discharge. 

Example.  What  is  the  pull  on  a  pump  rod,  when  dia. 
of  bucket  is  5  in.  and  water  is  raised  24  feet? 

Solution.     By  formula: 


P  =  H  AW=  &4Xr    —  X  62.5=204-45  Ibs. 
144 

which  is  the  pull  on  rod  necessary  to  operate  the  pump;  to 
this  must  be  added  the  amount  of  power  required  to  overcome 
friction  in  the  moving  parts. 

The  steam  pump  is  often  used  to  supply  the  feed  water 
to  the  boiler.  The  pump  in  such  cases  is  usually  made  with 
a  steam  piston  at  one  end  of  a  connecting  rod  and  the  water 
piston  at  the  other  end.  In  this  case  the  steam  piston  must 
be  enough  larger  in  diameter  than  the  water  piston  to  over- 
come the  friction  of  the  mechanism  and  leakage  in  the  valves, 
besides  the  steam  pressure  in  boiler,  against  which  the  pump 
is  working.  From  5%  to  40%  is  allowed  according  to  the 
condition  of  the  mechanism  of  pump. 


SHOP  MATHEMATICS  165 

TO  FIND  THE  CAPACITY  OF  A  PUMP  PER  HOUR 

RULE.  Find  the  cubical  contents  of  the  water  cylinder  per 
stroke  in  cubic  inches,  multiply  by  number  of  strokes  per  hour 
and  divide  the  product  by  231  to  find  the  number  of  gallons 
or  by  1,728  to  find  the  capacity  hi  cubic  feet. 


TO  FIND  THE  H.  P.  REQUIRED  TO  PUMP  WATER  TO  A 
GIVEN  HEIGHT 

RULE.  Multiply  the  weight  in  pounds  of  water  to  be 
raised  per  minute  by  the  height  hi  feet  and  divide  by  33,000; 
the  quotient  will  be  the  H.  P.  required. 

The  formula  is  *.  P. 

Example.  What  is  the  capacity  per  hour  of  a  pump  with 
water  piston  6  in.  dia.  and  8  in.  stroke,  when  the  piston 
makes  75  strokes  per  minute? 

Solution.  The  contents  of  water  cylinder,  if  cylinder  is 
filled  at  each  stroke  is  AxL  =  28.274X8  =  226.2  cu.  in. 

At  75  strokes  per  minute  there  will  be  75X60  =  4,500 
strokes  per  hour. 

If  the  piston  pumps  226.2  cu.  in.  per  stroke  then  for  one 
hour  it  will  pump 

226.2X4,500  =  1,017,900   cu.   in., 
and  1,017,900  +  331  =  4,406.4  gal-  per  hr. 

Example.  Find  the  H.  P.  required  to  pump  the  water 
of  above  example  to  a  height  of  50  ft.  above  source  of  supply. 

Solution.  If  a  pump  will  raise  4,406.33  gal.  of  water  per 
hour,  it  will  raise  4, 406. 33 +  60  =7 3.. $8  gal.  per  minute  and 
73.438  gal.  weighs  73. 438X8$  =  61 1.983  Ibs.  This  weight  of 


166  SHOP  MATHEMATICS 

water   is   to   be   pumped   50  ft.   high  per  minute;  then  by 
formula 


=       .=.^ 

33,000         33,000  660 


PROBLEMS 

1.  Find  the  number  of  gal.  of  water  delivered  per  hour 
by  a  single  action  pump  6  in.  dia.  with  a  12  in.  stroke  of  the 
water  piston  at  100  strokes  per  minute. 

2.  What  H.  P.  would  be  required  to  operate  the  pump 
of  problem  1  if  the  discharge  is  80  ft.  above  the  source  of 
supply? 

3.  How  many  gals,  per  hour  will  be  delivered  by  a  6  in. 
dia.  water  piston  with  10  in.  length  of  stroke  and  making  one 
stroke  per  second,  if  the  cylinder  only  fills  three-fourths  full 
at  each  stroke? 

4.  If  the  plunger  of  a  hydraulic  elevator  is  16  in.  dia. 
and  the  pressure  of  supply  pipe  is  120  Ibs.  per  sq.  in.,  what 
weight  can  be  lifted  if  car  and  plunger  weigh  400  Ibs.  and 
the  friction  of  moving  parts  takes  5%  of  the  power? 

5.  How  many  cu.  ft.  of   water  per  hour  will  a  single 
action  pump  12  in.  dia.,  15  in.  stroke  deliver  at  50  double 
strokes  per  minute,  when  water  cylinder  is  three-fourths  full 
per  stroke? 

6.  What  H.  P.  will  be  required  to  raise  the  water  of 
problem  5,  40  ft.  above  source  to  a  delivery  tank? 

7.  Find  the  amount  of  water  in  gals,  that  will  be  pumped 
with  a  4  in.  dia.  single  action  pump  with  water  piston  moving 
at  100  F.  P.  M.  and  cylinder  drawing   only    seven-eighths 
full  per  stroke. 


SHOP  MATHEMATICS  167 

8.  What  power  will  be  required  to  pump  water  200  ft. 
above  source  with  a  12  in.  dia.  double  action  water  cylinder 
moving  100  F.  P.  M.f 

9.  Find  the  power  required  to  pump  from  a  river  to  a 
tank  at  the  top  of  a  factory  building  67  ft.  from  river  to  top 
of  tank  with  an  8  in.  dia.  single  action  water  piston  moving 
100  F.  P.  M.f 

10.  What  steam  pressure  will  be  required  on  the  steam 
piston  of  a  direct  acting  pump  to  deliver  water  70  ft.  above 
source,  when  the  steam  and  water  pistons  are  both  8  in. 
dia.  if  25%  is  allowed  for  leakage  and  friction  in  the  mech- 
anism? 

11.  A  tank  10  ft.  dia.  and  10  ft.  high  is  set  on  the  top  of 
a  building  70  ft.  above  the  level  of  a  water  supply  from 
which  an  8  in.  dia.  single  action  piston  pump  moving  100 
ft.  per  minute  is  delivering  to  tank  at  200  F.  P.  M.    How 
long  will  it  take  to  fill  the  tank  if  30%  is  allowed  for  loss  by 
leakage,  etc.? 

Note.  The  velocity  of  water  through  a  pipe  at  200  F.  P.  M. 
requires  a  pipe  to  be  of  diameter  .J5V gal.  per  minute.  Between 
100  and  200  F.  P.  M.,  the  pipe  should  be  of  a  diameter  — 

I     gal,  per  minute 
\  velocity  in  F.  P.  M. 

12.  What  size  pipe  will  be  required  for  delivery  pipe  of 
problem  11? 

13.  When  water  is  pumped  through  a  delivery  pipe  at  150 
F.  P.  M.,  what  dia.  of  pipe  will  be  required  from  a  12  in.  dia. 
by  18  in.  stroke  piston  of  a  single  action  pump  making  60 
strokes  per  minute? 

14.  When  the  piston  travel  is  125  F.  P.  M.  what  is  the 
number  of  cu.  ft.  of  water  pumped  per  minute  by  a  double 
action  pump  with  10  in.  dia.  of  piston? 


168  SHOP  MATHEMATICS 

15.  What  is  the  H.  P.  required  to  run  the  pump  of  prob- 
lem 14,  when  it  delivers  to  a  tank  72  ft.  above  the  source? 

16.  Find  the  size  of  a  delivery  pipe  required  for  problem 
15  if  it  were  to  discharge  into  the  tank  at  150  F.  P.  M. 

17.  When  the  bucket  of  a  hand  suction  pump  is  3  in. 
dia.  and  the  supply  is  drawn  from  a  depth  of  30  ft.,  what 
pressure  is  required  on  the  handle  26  in.  from  the  fulcrum 
when  the  center  of  bucket  rod  is  5  in.  from  the  fulcrum? 

18.  When  the  piston  travel  of  a  double  action  steam 
pump  is  150  F.  P.  M.  and  15  in.  dia.,  find  the  H.  P.  required 
to  fill  a  supply  tank  62  £  ft.  above  the  source. 

19.  What  dia.  of  pipe  will  be  required  to  deliver  the  water 
of  problem  18  at  a  speed  of  200  F.  P.  M.?     • 


STANDARD  UNITS  OF  WEIGHTS  AND 
MEASURES 

Measure  is  that  by  which  extension,  capacity,  force, 
duration,  or  value  is  estimated  or  determined. 

The  weight  of  a  body  is  the  measure  of  the  force  of  the 
earth's  attraction  for  that  body,  commonly  called  the  force 
of  gravity. 

Linear  or  Long  measure  is  used  in  measuring  distances. 
Table  of  Linear  Measure 

12    inches  =1  foot  (ft.). 

3    feet  =1  yard  (yd.). 

5*  yds.  or  16*  ft.=  1  rod  (rd.). 
320    rods  =  1  mile. 

1    mile  =320    rds.=  1,760    yds.  =  5,280    ft.= 

63,360  in. 

Surface  measure  is  used  in  measuring  in  two  dimensions, 
length  and  breadth. 

Table  of  Square  Measure 

144    square  inches  =  1  square  foot. 

9    square  feet  =  1  square  yard. 

30^  sq.  yds.  or  272£  sq.  ft.  =  1  square  rod. 
160    square  rods  =  1  acre. 

640    acres  =  1  square  mile. 

1    sq.  rd.  =30^  sq.  yds.  =  272^  sq.  ft.  = 

39,204  sq.  in. 

Pressures  of  liquids  and  gases  are  usually  given  in  pounds 
per  square  inch  or  square  foot. 


170  SHOP  MATHEMATICS 

Cubic  measure  is  used  in  measuring  the  volume  of  solids. 
Table  of  Solid  or  Cubic  Measure 

1,728  cubic  inches  =  1  cubic  foot. 
27  cubic  feet     =  1  cubic  yard. 
1  cu.  yd.          =  27  cu.  ft.  =  46,656  cu.  in. 

The  measure  of  volume  for  liquids. 

Table  of  Liquid  Measure 

4    gills      =  1  pint. 
2    pints     =  1  quart. 
4    quarts  =  1  gallon. 

31^  gallons  =  1  barrel. 

Liquids  are  sometimes  measured  with  cubic  measure,  the 
U.  S.  gal.  contains  231  cu.  in.  so  that  a  cu.  ft.  of  water  con- 
tains approximately  1\  gallons. 

One  gallon  of  water  taken  at  the  temperature  of  maximum 
density,  39.1°F.,  weighs  8.3389  Ibs.,  avoirdupois,  which  is 
approximately  8J  Ibs. 

A  cubic  foot  of  water  weighs  62.355  Ibs.,  approximately 
62  £  Ibs.  Water  freezes  at  32°  F.  or  0°  C.,  and  boils  at  212°  F 
(Fahrenheit)  or  100°  C.  (Centigrade). 

The  standard  unit  for  weight  in  shop  practice  is  the 
avoirdupois  pound. 

Table  of  Avoirdupois  Weight 

16  ounces  =  1  pound. 
100  pounds  =  1  hundred  weight  (cwt.). 
20  cwt.      =  1  ton. 
1  ton        =  20  cwt.  =  2,000  Ibs.  =  32,000  oz. 


SHOP  MATHEMATICS  171 

Time  is  the  measure  of  duration. 

Table  of  Measure  for  Time 

60  seconds  =  1  minute. 
60  minutes  =  1  hour. 
24  hours     =  1  day. 
7  days       =  1  week. 

365  days       =  1  year. 

366  days  on  leap  years. 

Angular  measure  is  used  in  measuring  angles. 
Table  of  Angular  Measure 

60  seconds  (60")     =  1  minute  (I'). 
60  minutes  (60')    =  1  degree  (1°). 
360  degrees  (360°)  =  1  circumference. 

Money  is  the  measure  for  values.    It  is  used  as  a  medium 
of  exchange. 

Table  of  U.  S.  Money 

10  mills    =  1  cent. 
10  cents    =  1  dime. 
10  dimes  =  1  dollar. 
10  dollars  =  1  eagle. 
1  eagle    =  $  1 0  =  100  dimes  =  1 ,000  cents  =  10,000  mills. 

MISCELLANEOUS  MEASURES 
Table 

12  units  =  1  dozen. 
12  dozen  =  1  gross. 
12  gross  =  1  great  gross. 
1  gross  =  12  doz.=  144  units. 


172  SHOP  MATHEMATICS 

Weights  of  Materials 

1  cu.  in.  of  cast  iron  weighs  .2604  Ib.  melts  at  2,500°  F. 

1  cu.  in.  of  wr't.  iron  weighs  .2779  Ib.  melts  at  3,100°  F. 

1  cu.  in.  of  steel  weighs  .2833  Ib.  melts  at  3,000°  F. 

1  cu.  in.  of  copper  weighs  .3195  Ib.  melts  at  1 ,930°  F. 

1  cu.  in.  of  brass  ]    .  [  weighs  .3029  Ib.  melts  at 

(  zinc  35 ) 

1  cu.  in.  of  lead  weighs  .4106  Ib.  melts  at     625°  F. 

1  cu.  in.  of  aluminum          weighs  .0963  Ib.  melts  at  1,160°  F. 
The  approximate  weight  of  round  bar  iron  or  steel  can  be 
found  by  the  formula 
T_(dX4)2 
6      ' 

where  L  =  weight  of  one  foot  in  length  of  the  bar 
and        d  =  diameter  of  the  bar  in  inches. 

Horse  Power 

The  horse  power  is  an  expression  for  foot-pounds  of  useful 
work  accomplished  in  a  specified  time.  1  H.  P.  =  33, 000 
pounds  raised  1  foot  in  one  minute,  or  1  pound  raised  33,000 
feet  in  one  minute. 

Electric  Units 

The  volt  is  the  unit  of  electrical  pressure. 

The  ampere  is  the  unit  of  current  strength  or  rate  of  flow. 

The  ohm  is  the  unit  of  resistance. 

The  watt  is  the  unit  of  power. 

The  ohm,  ampere  and  volt  are  defined  in  terms  of  one  an- 
other as  follows: 

Ohm,  the  resistance  of  a  conductor  through  which  a  current 
of  one  ampere  will  pass  when  the  electro-motive  force  is  one 
volt. 


SHOP  MATHEMATICS  173 

Ampere,  the  quantity  of  current  which  will  flow  through 
a  resistance  of  one  ohm  when  the  electro-motive  force  is  one 
volt. 

Volt,  the  electro-motive  force  required  to  cause  a  current 
of  one  ampere  to  flow  through  a  resistance  of  one  ohm. 

The  relation  which  these  quantities  bear  to  one  another  is 
expressed  by  Ohm's  Law. 

E  M  F  in  volts 

Current  in  amperes  =  -=: — :— : j — . 

Resistance  in  ohms 

E 

Then         C-I-. 
H 

where        E  =  the  electro-motive  force  in  volts, 
R  =  resistance  in  ohms, 
C=  current  in  amperes. 
The  electric  H.  P.  =  746  watts. 

METRIC  SYSTEM  OF  MEASURES 

The  metric  system  is  used  in  almost  all  the  European 
countries  and  is  authorized  by  law  in  the  United  States  and 
is  in  general  use  for  all  scientific  treatises. 

Metric  Linear  Measure 

10  millimeters  (mm.)  =  l  centimeter  (cm.). 

10  centimeters  =1  decimeter  (dm.). 

10  decimeters  =  1  meter  (m.). 

10  meters  =1  dekameter  (Dm.). 

10  dekameters  =  1  hektometer  (Hm.). 

10  hektometers  =  1  kilometer  (Km.). 

10  kilometers  =  1  myriameter  (Mm.) . 

i  Mm.  =  10  Km.  =  100  Hm.  =  1,000  Dm.  =  10,000  meters. 
1     m.=  10  dm.  =  100    cm.  =  1,000  mm.  =  39.37  inches. 
The  unit  for  metric  linear  measure  is  the  meter. 


174  SHOP  MATHEMATICS 

Metric  Square  Measure 

100  sq.  millimeters    =  1  sq.  centimeter  (sq.  cm.). 
100  sq.  centimeters   =  1  sq.  decimeter  (sq.  dm.) 
100  sq.  decimeters     =1  sq.  meter  (sq.  m.). 
100  sq.  meters  =  1  sq.  dekameter  (sq.  Dm.). 

100  sq.  dekameters   =  1  sq.  hektometer  (sq.  Hm.). 
100  sq.  hektometers  =  1  sq.  kilometer  (sq.  Km.). 
The  unit  for  metric  square  measure  is  the  square  meter  for 
small  surfaces,  and  the  are  for  land  measurements. 
100  centares=  1  are        1  sq.  meter  =  1.196  sq.  yard 

100  ares=  1  hectare 


Metric  Cubic  Measure 

1,000  cu.  millimeters  =  1  cu.  centimeter  (cu.  cm.). 
1,000  cu.  centimeters  =  1  cu.  decimeter  (cu.  dm.). 
1,000  cu.  decimeters    =  1  cu.  meter  (cu.  M.). 
The  unit  for  cubic  measure  is  the  cubic  meter  and  1  cu.  m. 
=  1.308  cu.  yard. 

Metric  Measure  of  Capacity 

10  milliliters     =  1  centiliter  (cl.). 

10  centiliters    =1  deciliter  (dl.). 

10  deciliters      =1  liter  (1.). 

10  liters  =  1  dekaliter  (DL). 

10  dekaliters    =  1  hektoliter  (HI). 

10  hektoliters  =  1  kiloliter  (Kl.). 

The  unit  for  capacity  for  both  liquid  and  dry  measure  is 
the  liter. 
1  liter  =  .908  of  a  dry  quart.     1  liter  =  1.0567  of  a  liquid  quart. 


SHOP  MATHEMATICS 


175 


Metric  Measure  of  Weight 

10  milligrams     =1  centigram  (eg.). 
10  centigrams    =1  decigram  (dg.). 
10  decigrams      =  1  gram  (g.). 
10  grams  =1  dekagram  (Dg.). 

10  dekagrams    =  1  hektogram  (Hg.). 
10  hektograms  =  1  kilogram  (Kg.). 
10  kilograms      =1  myriagram  (Mg.). 
10  myriagrams  =1  quintal  (Q.). 
10  quintals         =  1  tonneau  (T.). 

The  unit  of  weight  is  the  gram.      1   gram  =.03527  oz. 
Avoirdupois. 

EQUIVALENTS  OF  COMMON  AND  METRIC  SYSTEMS 


Linear  Measure 


1  inch  =  2.54  cm. 
1  foot  =0.3048  m. 
1  yard  =  0.9 144m. 
1  rod    =  5.029  m. 
1  mile  =1.6093  Km. 


1  cm.  =  0.3937  in. 
1  dm.  =0.328  ft. 
1m.     =1.0936  yds. 
1  Dm.=  1.9884rds. 
1  Km.  =  0.62137  mile 


Square  Measure 


1  sq.  in.    =   6.452  sq.  cm. 
1  sq.  ft.     =   0.0929  sq.  m. 
1  sq.  yd.   =   0.8361  sq.  m. 
1  sq.  rd.    =25.293  sq.  m. 
1  acre       =40.47  ares 
1  sq.  mile  =  259  hectares 


1  sq.  cm.  =0.155  sq.  in. 
1  sq.  dm.  =  0.1076  sq.  ft. 
1  sq.  m.     =  1.196  sq.  yds. 
1  are         =3.954  sq.  rds. 
1  hectare  =2.471  acres 
1  sq.  Km.  =  0.3861  sq.  mile 


176 


SHOP  MATHEMATICS 
Cubic  Measure 


1  cu.  in.    =  16.387  cu.  cm. 
1  cu.  ft.    =28.317  cu.  dm. 
1  cu.  yd.  =0.7645  cu.  m. 
1  cord       =  3.624  steres 


1  cu.  cm.  =0.061  cu.  in. 
1  cu.  dm.  =0.0353  cu.  ft. 
1  cu.  m.    =  1.308  cu.  yds. 
1  stere      =0.2759  cord 


Measures  of  Capacity 


1  liquid  qt.  =  0.9463  liters 
1  dry  qt.      =1.101  liters 
1  liquid  gal.  =  0.3785  Dl. 
1  peck  =  0.881  Dl. 

1  bushel       =0.3524  HI. 


1  liter  =  1.0567  liquid  qts. 
1  liter  =  0.908  dry  qts. 
1  Dl.  =  2.6417  liquid  gals. 
1  Dl.  =1.135  pecks 
1  HI.  =  2.8375  bushels 


Measures  of  Weight 


1  grain  Troy=   0.0648  gram. 

1  oz.  Avoir.  =  28.35  g. 

1  oz.  Troy     =31.104g. 

1  Ib.  Avoir.   =   0.4536  Kg. 

1  Ib.  Troy     =  0.3732  Kg. 

1  ton  (2000  lbs.)=  0.9072  tonneau, 

1  tonneau  =1.1023  tons  (2000  Ibs.) 


1  g.     =   0.03527  oz.  Avoir. 
1  g.     =   0.03215  oz.  Troy 
1  g.     =  15.432  grains  Troy 
1  Kg.  =   2.2046  Ibs.  Avoir. 
1  Kg.=   2.679  Ibs.  Troy 


SHOP  MATHEMATICS 


177 


DECIMAL  EQUIVALENTS  OF  PARTS  OF  AN  INCH 


A  .01563 

A  .03125 

A  .04688 

A  .0625 

A  .07813 

A  .09375 

A  .10938 

i  .125 

A  .14063 

A  .15625 

Ji  .17188 

A  .1875 

H  .20313 

A  .21875 

if  .23438 

J  .25 

H  .26563 

A  .28125 

H  .29688 

A  -3125 

ft  .32813 

H  .34375 

ft  .35938 

I  .375 

H  .39063 

Jf  .40625 

tt  .42188 

A  -4375 

H  .45313 

H  .46875 

ft  .48438 

*  .5 


H  .51563 

J}  .53125 

H  .54688 

A  -5625 

ft  .57813 

H  .59375 

H  .60938 

§  .625 

H  .64063 

ft  .65625 

If  .67188 

ft  .6875 

H  .70313 

H  .71875 

tf  .73438 

I  .75 

tt  .76563 

H  .78125 

ft  .79688 

H  .8125 

ft  .82813 

Ji  .84375 

if  .85938 

t  .875 

H  .89063 

H  .90625 

H  .92188 

tt  .9375 

tt  .95313 

H  .96875 

ft  .98438 

1  1.00000 


USE  OF  FORMULAS 

A  formula,  as  used  in  technical  books  and  papers,  may  be 
defined  as  a  rule  in  which  letters  and  symbols  are  used  in 
place  of  numbers  or  words.  A  formula,  then,  is  a  short  way 
of  expressing  a  rule,  and  is  more  convenient,  as  it  shows  at  a 
glance  the  operations  to  be  performed. 

For  example,  the  volume  of  a  rectangular  solid  is  equal  to 
the  product  of  its  three  dimensions,  length,  width,  and  height. 

The  same  statement  would  be  expressed  by  the  formula, 
V  =  LWH. 

The  signs  of  addition,  subtraction,  multiplication  and 
division  are  used  in  formulas  in  the  same  way  as  in  arithmetic 
except  that  the  sign  of  multiplication  between  two  letters  is 
usually  omitted ;  as  V  =  LWH  means  V  =  L  X  W  X  H. 

In  working  out  or  solving  a  formula,  the  numerical  values 
of  all  but  one  of  the  letters  must  usually  be  known.  These 
values  are  substituted  for  their  respective  letters  and  the 
value  of  the  remaining  letter  may  then  be  easily  found. 

Example.     In  the  formula  S=%  gt2.     Let  g  =  32,  t=5. 
Find  S. 
Solution.     By  formula,  S=$  of  32X5*  =  16X25  =  400. 

Every  formula  is  in  reality  an  algebraic  equation,  and  a 
knowledge  of  some  of  the  simpler  principles  of  equations  will 
be  of  great  assistance  in  working  out  formulas. 

The  equation  may  be  illustrated  by  reference  to  a  scale 
pan  with  equal  arms.  If  equal  weights,  say  five  pounds,  are 
put  in  each  pan,  they  will  just  balance.  If  two  pounds  is 


SHOP  MATHEMATICS 


179 


t 


3 


added  to  one  pan,  two 
pounds  must  be  added 
to  the  other  pan,  and  if 
three  pounds  is  taken 
from  one  pan,  three 
pounds  must  also  be 
taken  from  the  other  or 
the  pans  will  not  balance. 

The  same  is  true  of  the 
equation;  whatever  oper- 
ation is  performed  on  one  side,  the  same  operation  must 
be  performed  on  the  other  side. 

The  following  operations  may  be  performed  on  an  equation 
without  changing  its  value. 

I  The  same  number  may  be  added  to  each  side. 

II  The  same  number  may  be  subtracted  from  each  side. 

III  Each  side  may  be  multiplied  by  the  same  number. 

IV  Each  side  may  be  divided  by  the  same  number. 
V  The  same  root  may  be  taken  of  each  side. 

VI    Each  side  may  be  raised  to  the  same  power. 

TRANSPOSING  TERMS 

Example.  Five  added  to  five  times  a  certain  number  is 
equal  to  15.  What  is  the  number? 

Solution.  Let  x  =  the  unknown  number.  Then  the  state- 
ment of  the  equation  will  be  5x-\-5  =  15. 

But  since  by  (II)  5  may  be  taken  from  each  side,  the  state- 
ment then  becomes,  5x  =  10. 

By  (IV)  x=2. 

In  the  above  example  the  statement: 


180  SHOP  MATHEMATICS 

5x-\-  5  =  15  could  be  changed  to  read  as  follows: 

5x  =  15—  5  from  which  could  be  obtained  the  following. 

RULE :  Any  term  of  an  equation  can  be  moved  from  one 
side  to  the  other  by  changing  its  sign. 

This  is  called  transposing  a  term. 

CLEARING  OF  FRACTIONS 

Example.  The  sum  of  the  half  and  the  fifth  parts  of  a 
number  is  7.  What  is  the  number? 

Solution.    Let  z  =  the  unknown  number. 
Then  the  statement  will  be  as  follows: 

-+-  =  7 
2^5     '• 

By  (III)  it  is  possible  to  multiply  both  sides  of  the  equa- 
tion by  the  same  number,  in  this  case  10,  which  is  the  L.C.M . 
of  2  and  5. 

Then  the  equation  becomes: 

5x+2x  =  70 
7x  =  70 

By  (IV)  x  =  10 

To  prove  that  the  result  is  correct,  the  value  found  for  x 
when  substituted  in  the  original  equation  should  make  both 
sides  identical. 

Then  10  .10 

y+y  =  7,  or  7  =  7. 

This  is  called  checking  the  problem. 

A  coefficient  is  any  factor  of  the  rest  of  the  term;  as,  in  ab, 
a  is  the  coefficient  of  6  and  6  is  the  coefficient  of  a,  but  in  an 
algebraic  formula  the  coefficient  is  usually  understood  to 


SHOP  MATHEMATICS  181 

mean  the  numerical  coefficient  of  the  given  term;  as,  5  is  the 
coefficient  of  x  in  the  expression  5x. 

Terms  that  differ  only  in  their  numerical  coefficients  are 
called  similar  and  may  be  added  or  subtracted  by  finding  the 
sum  or  difference  of  then*  coefficients  and  affixing  to  the 
result  the  common  letters. 

Examples.        7a+3a+6a  =  16a. 

2xm-\-12xm—8xm  =  6xm. 
8abc  —  Sabc  =  5abc. 

The  law  for  precedence  of  signs.  When  the  signs  +,  —  ,  X, 
and  -v-  occur  in  the  same  formula,  the  operations  of  multipli- 
cation and  division  must  be  performed  before  those  of  addi- 
tion and  subtraction;  as, 

9X3—6  +  2  =  27-3  =  24. 

The  law  of  signs  for  addition.  In  adding  similar  terms 
having  +  and  —  signs,  subtract  the  less  number  from  the 
greater  (arithmetically)  and  prefix  the  sign  of  the  greater 
number;  as,  5x+(—  8x)  =  —  3x. 

The  law  of  signs  for  subtraction.  Change  the  sign  of  every 
term  in  the  subtrahend  and  then  proceed  as  in  addition;  as, 
5x-  (-8x)  =  5x+(+8x)  =  ISx. 

The  law  of  signs  for  multiplication.    The  product  of  two 
terms  of  like  signs  has  the  plus  sign;    the  product  of  two 
terms  of  unlike  signs  has  the  minus  sign;  as, 
(+a)X(+6)=a6. 


(-a)X(-b)=ab. 


An  exponent  denotes  the  number  of  times  a  quantity  is 
used  as  a  factor;    as,  aXaXa  =  a3. 


182  SHOP  MATHEMATICS 

Then  multiplying  a2  by  a3  means,  (aXa)X(aXaXa)  or  a 
used  as  a  factor  five  times.  The  result  is  written  a5. 

From  the  above  follows: 

The  law  for  exponents  in  multiplication.  The  exponent  of 
a  letter  in  the  product  is  equal  to  the  sum  of  its  exponents  in 
the  factors. 

The  law  of  signs  for  division.  If  the  numbers  to  be  divided 
have  the  same  sign,  the  sign  of  the  quotient  will  be  plus;  if 
the  numbers  to  be  divided  have  unlike  signs  the  sign  of  the 
quotient  will  be  minus;  as, 


The  law  for  exponents  in  division.  The  exponent  of  a  letter 
in  the  quotient  is  equal  to  its  exponent  in  the  dividend  minus 
its  exponent  in  the  divisor. 

Example.  Divide  a4  by  a2. 

c,  ,  ..  a4    aXaXaXa     rp        f  c    ^ 

Solution.  —=  —  —.     Two    factors    of    the 

a-          aXa 

dividend  will  be  canceled  by  the  two  factors  of  the  divisor 
and  the  quotient  =  a2,  or  a4-r-a2  =  a4~2=a2. 


SHOP  MATHEMATICS  183 

PROBLEMS  INVOLVING  USE  OF  FORMULAS 

Example.  Find  the  value  of  V  in  the  formula  V=LWH. 
When  L=20,  W=15  and  H  =  5. 

Solution.  Substituting  the  known  values  for  the  letters 
L,  W,  and  H  in  the  formula: 

Then  V=  20X1  5X5  =  1500. 

Example.  What  is  the  value  for  H  in  the  formula  V= 
L  W  H,  when  V=SOOO,  L  =  45  and  W=20? 

Solution.  Substituting  the  known  values  for  the  letters 
V,  L,  and  W  in  the  formula: 

Then  3000  =  45X20XH. 

The  quantities  L,  W  and  H  are  factors  of  the  quantity  V. 
Therefore  the  values  of  one  of  the  factors  of  V  must  be  the 
quotient  obtained  by  dividing  V  by  the  product  of  the  other 
factors; 

Then  by  transposing,  B- 


In  like  manner  any  other  term  can  be  found,  as, 

V 


WH  LH 

E 

1.  G  =  —~.     Find  E  when  C  =  10  and  R  =  5. 

ti 

2.  j^  =  ~-     FindF  when  D  =  10  and  L»=#0. 

3.  7  =  f  X£.     Find  £  when  V=816  and  H=^. 

o 

4.  — =4     Find  C  when  P  =  6,  W  =  96  and  L  =  7. 


184  SHOP  MATHEMATICS 

5.  PXa  =  WXb.    Find  the  value  of  each  letter  in  terms 
of  the  other  letters. 

a      c,.  a.     Find  a  in  terms  of  c  and  sin  A.    Also  find 

6.  SmA=—  .  ,         ,    .     . 

c          cm  terms  of  a  and  sin  A. 

7.  Tan  A  =—-   Find  tan  A  when  a  =  8  and  b  =  6. 

o 

8.  C  =  ^-(F°-32°}.    Find  F°  in  terms  of  C°. 

t7 

Note.  In  the  solution  of  this  problem  the  result  could  be 
obtained  by  first  taking  f  of  T''0  and  from  this  product  sub- 
tracting f  of  32°;  but  the  parenthesis  indicates  that  the 
enclosed  terms  are  to  be  treated  as  one;  32°  should  first  be 
subtracted  from  ^°  and  then  f  of  the  difference  should  be 
found. 

g  (p°—g2°} 
From  the  formula  C°  =  —    — ^— ^  -  a  formula  for  F°  may 

be  derived  as  follows: 

Clearing  of  fractions,  9C°  =  5  (F°  -  32°) 

Removing  parenthesis,         9C° = 5F°  - 160° 
Transposing,  5F°=9C°+160° 

Dividing  by  the  coefficient  )  no_9C°  .  _-0 

of  the  unknown  term,  (  5 

The  operation  of  removing  the  parenthesis  from  a  formula 
should  be  performed  first  and  the  resulting  number  treated 
according  to  the  signs  of  operation  that  affect  the  whole 
parenthesis;  thus,  |  (3+5)2  =  f  (5)2  = 

Again  6a-(3b-2c+l}=6a-Sb+2c—l. 
Also,  6a+(3b-2c+l)=6a+Sb-2c+l. 
From  the  above  it  follows:  that 


SHOP  MATHEMATICS  185 

To  remove  a  parenthesis.  When  the  plus  sign  precedes  it, 
the  signs  of  all  the  terms  within  the  parenthesis  remain  un- 
changed. But  if  the  minus  sign  precedes,  the  sign  of  every 
term  within  the  parenthesis  must  be  changed;  this  is  but  an 
illustration  of  the  law  of  signs  in  subtraction  where  the  signs 
of  all  the  terms  in  the  subtrahend  are  changed  before  adding. 

When  a  parenthesis  or  other  sign  of  union  occurs  within  a 
parenthesis,  the  inner  ones  are  usually  removed  first;  as, 


5a+8a+4a+lla+9a= 
9.     Change  77°  F  to  the  equivalent  Centigrade  reading. 

10.  Change  18°  F  to  the  equivalent  Centigrade  reading. 

11.  Change  10°  C  to  the  equivalent  Fahrenheit  reading. 

12.  Change  —  10°  C  to  the  equivalent  Fahrenheit  reading. 


186  SHOP  MATHEMATICS 

SIMPLE  TRIGONOMETRIC  FUNCTIONS 

The  right  triangle  ABC  has  six  parts 
or  elements;  three  sides  a,  b,  c,  and 
three  angles  A,  B,  and  (7.  When  three 
of  these  elements  are  given  one  of  the 
known  parts  being  a  side  the  three 
unknown  parts  can  be  determined  by  computation. 

The  process  of  finding  the  unknown  parts  is  called  solving 
the  triangle. 

The  radios  of  the  sides  of  the  right  triangle  are   called 
trigonometric  functions  and  are  named  as  follows: 

The  sine  of  an  angle  is  the  ratio  of  the  side  opposite  the 

angle  to  the  hypotenuse,  as  sin  A=  — ,  sin  B  =  — . 

C  C 

The  cosine  of  an  angle  is  the  ratio  of  the  adjacent  side  to 
the  hypotenuse,  as  cos  A  =  — ,  cos  B  =  — . 

C  C 

The  tangent  of  an  angle  is  the  ratio  of  the  side  opposite 

to  the  side  adjacent,  as  tan  A  —  -J-,  tan  B  =  — . 

o  a 

The  cotangent  of  an  angle  is  the  ratio  of  the  adjacent  side 

to  the  opposite  side,  as  cot  A  =  — ,   cot  B=-^~. 

a  o 

The  secant  of  an  angle  is  the  ratio  of  the  hypotenuse  to 

C  C 

the  adjacent  side,  as  sec  A  =  -j--,  sec  B  =  — . 

The  cosecant  of  an  angle  is  the  ratio  of  the  hypotenuse 

C  C 

to  the  opposite  side,  as  esc  A  =  — ,  esc  5  =  -=-. 

a  b 

The  following  arrangement  of  the  functions  of  an  angle 
will  be  found  convenient,  and  with  the  tables  of  natural 


SHOP  MATHEMATICS 


187 


functions  given  on  pagi 
necessary  for  the  solutioi 

(1).     Sine 
(2).     Side  opposite 
(3)  .     Cosine 
(4).     Side  adjacent 
(5).     Tangent 
(6).     Side  opposite 
(7)  .     Cotangent 
(8).     Side  adjacent 
(9)  .     Hypotenuse 

(  1  rf\          TTTr»r»t  on  11  ae> 

3s  189  to  193  furnish  all  the  data 
i  of  right  triangles, 
side  opposite 

hypotenuse 
=  hypotenuse  X  sine 
side  adjacent 

hypotenuse 
=  hypotenuse  X  cosine 
side  opposite 

side  adjacent 
=  side  adjacent  X  tangent 
side  adjacent 

side  opposite 
=  cotangent  X  side  opposite 
side  opposite 

sine 
side  adjacent 

How  to  Use  the  Tables 

Example.     If  a  =  47  ft.,  and  c  =  63  ft.,  find  angle  A. 

Solution.     By  proposition  (1)  sin  A  =  —  =  —  =  .74603.  Re- 

C       oo 

fer  to  page  193  in  column  marked  sines  at  bottom  on  right, 
find  .74606  just  above  the  figures  48  in  the  column  marked 
D  for  degrees  and  on  the  same  line  with  figures  15  in  column 
marked  M  for  minutes.  It  is  evident  that  the  angle  A  is  very 
close  to  4$°  15',  as  the  next  smaller  angle  given  in  the  table 
is  48°  0'  for  which  the  sine  is  .74314  and  as  the  machinists' 
protractor  has  quarter  degrees  for  the  smallest  graduations 


188  SHOP  MATHEMATICS 

the  nearest  approximation  is  taken  to  the  size  of  the  angle 
as  found  in  the  table,  that  is  48°  15' . 

Example.     Find  cosine  of  angle  A  when  A  —  35°  30' . 

Solution.  On  page  192  in  column  headed  D  on  left  side, 
follow  down  the  column  until  the  figure  35  is  reached,  in  the 
next  column  headed  M  on  the  same  line  is  o,  under  this 
is  15  and  under  15  is  30  which  is  the  angle  required,  that  is, 
35°  30' .  On  this  line  at  the  right  under  column  headed 
cosines  (at  top)  will  be  found  the  decimal  .81412  which  is  the 
cosine  of  angle  35°  3Cf . 

Note.  For  angles  between  0°  and  45°  the  names  of  the  functions 
are  found  at  the  top  of  the  page  and  the  angles  at  the  left;  for  all 
angles  between  45°  and  90°  the  names  of  the  functions  are  found 
at  the  bottom  of  the  page  and  the  angles  at  the  right. 


SHOP  MATHEMATICS  189 

NATURAL  TRIGONOMETRICAL  FUNCTIONS 


D 

M 

Sines 

Cosines 

Tangents 

Cotangents 

Secants 

Cosecants 

0 

0 

.00000 

1.0000 

.00000 

Infinite 

I.  0000 

Infinite 

90 

0 

15 

.00436 

.99999 

.00436 

229  .  182 

1.0000 

229  .  18 

45 

30 

.00873 

.99996 

.00873 

114.589 

1.0000 

114.59 

30 

45 

.01309 

.99991 

.01309 

76.3900 

1.0001 

76.397 

15 

1 

0 

.01745 

.99985 

.01746 

57.2900 

1.0001 

57.299 

89 

0 

15 

.02181 

.99976 

.02182 

45.8294 

1.0002 

45.840 

45 

30 

.02618 

.99966 

.02619 

38.1885 

1.0003 

38.202 

30 

45 

.03054 

.  99953 

.03055 

32.7303 

1.0005 

32  .  746 

15 

2 

0 

.03490 

.99939 

.03492 

28.6363 

1.0006 

28.654 

88 

0 

15 

.03926 

.99923 

.03929 

25.4517 

1.0008 

25.471 

45 

30 

.04362 

.99905 

.04366 

22.9038 

1.0009 

22.926 

30 

45 

.04798 

.99885 

.04803 

20.8188 

1.0011 

20.843 

15 

3 

0 

.05234 

.99863 

.05241 

19.0811 

1.0014 

19.107 

87 

0 

15 

.05669 

.99839 

.05678 

17.6106 

1.0016 

17.639 

45 

30 

.06105 

.99813 

.06116 

16.3499 

1.0019 

16.380 

30 

45 

.06540 

.99786 

.06554 

15.2571 

1.0021 

15.290 

15 

4 

0 

.06976 

.99756 

.06993 

14.3007 

1.0024 

14.336 

86 

0 

15 

.07411 

.99725 

.07431 

13.4566 

1.0028 

13.494 

45 

30 

.07846 

.99692 

.07870 

12.7062 

1.0031 

12.745 

30 

45 

.08281 

.99657 

.08309 

12.0346 

1.0034 

12.076 

15 

5 

0 

.08716 

.99619 

.08749 

11.4301    1.0038 

11.474 

85 

0 

15 

.09150 

.99580 

.09189 

10.8829    1.0042 

10.929 

45 

30 

.09585 

.99540 

.09629 

10.3854  !  1.0046 

10.433 

30 

45 

.10019 

.99497 

.  10069 

9.93101   1.0051 

9.9812 

15 

6 

0 

.  10453 

.99452 

.10510 

9.51436   1.0055 

9.5668 

84 

0 

15 

.  10887 

.99406 

.  10952 

9.13093!  1.0060 

9  .  1855 

45 

30 

.11320 

.99357 

.11394 

8.77689   1.0065 

8.8337 

30 

45 

.11754 

.99307 

.11836 

8.44896J  1.0070 

8.5079 

15 

7 

0 

.12187 

.99255 

.  12278 

8.14435;  1.0075 

8  .  2055 

83 

0 

15 

.12620 

.99200 

.12722 

7.86064:  1.0081 

7.9240 

45 

30 

.  13053 

.99144 

.13165 

7.59575!  1.0086 

7.6613 

30 

45 

.  13485 

.99087 

.  13609 

7.  34786  i  1.0092 

7.4156 

15 

8 

0 

.13917 

.99027 

.  14054 

7.  11537  |  1.0098 

7.1853 

82 

0 

15 

.  14349 

.98965 

.  14499 

6.89688;   1.0105 

6.9690 

45 

30 

.  14781 

.98902 

.  14945 

6.69116;  1.0111 

6  .  7655 

30 

45 

.15212 

.98836 

.  15391 

6.  49710  i  1.0118 

6.5736 

15 

9 

0 

.15643 

.98769 

.15838 

6.31375!   1.0125 

6.3924 

81 

0 

15 

.  16074 

.98700 

.  16286 

6.14023,  1.0132 

6.2211 

45 

30 

.16505 

.98629 

.16734 

5.975761  1.0139 

6.0589 

30 

45 

.  16935 

.98556 

.17183 

5.81966   1.0147 

5.9049 

15 

10 

0 

.17365 

.98481 

.17633 

5.67128!   1.0154 

5  .  7588 

80 

0 

Cosines 

Sines 

Cotangents 

Tangents  Cosecants 

Secants 

D 

M 

From  80°  to  90°  read  from  bottom  of  table  upwards. 


190  SHOP  MATHEMATICS 

NATURAL  TRIGONOMETRICAL  FUNCTIONS 


D 

M 

Sines 

Cosines 

Tangents 

Cotangents 

Secants 

Cos  ecants 

10 

0 

.17365 

.98481 

.17633 

5.67128 

1.0154 

5.7588 

80 

0 

15 

.17794 

.  98404 

.  18083 

5.53007 

1.0162 

5.6198 

I  45 

30 

.18224 

.98325 

.  18534 

5.39552 

1.0170 

5.4874 

30 

45 

.  18652 

.98245 

.  18986 

5.26715 

1.0179 

5.3612 

15 

11 

0 

.  19081 

.98163 

.19438 

5.14455 

1.0187 

5.2408 

79 

0 

15 

.  19509 

.98079 

.19891 

5.02734 

1.0196 

5.1258 

45 

30 

.  19937 

.97992 

.  20345 

4.91516 

1.0205 

5.0158 

30 

45 

.20364 

.97905 

.  20800 

4.80769 

1.0214 

4.9106 

15 

12 

0 

.20791 

.97815 

.21256 

4.70463 

1.0223 

4.8097 

78 

0 

15 

.21218 

.97723 

.21712 

4.60572 

1.0233 

4.7130 

45 

30 

.21644 

.97630 

.22169 

4.51071 

1.0243 

4.6202 

30 

45 

.22070 

.97534 

.  22628 

4.41936 

1.0253 

4.5311 

15 

13 

0 

.  22495 

.97437 

.  23087 

4.33148 

1.0263 

4.4454 

77 

0 

15 

.22920 

.97338 

.23547 

4.24685 

1.0273 

4.3630 

45 

30 

.  23345 

.97237 

.24008 

4.16530 

1.0284 

4.2837 

30 

45 

.23769 

.97134 

.24470 

4.08666 

1.0295 

4.2072 

15 

14 

0 

.24192 

.97030 

.  24933 

4.01078 

1.0306 

4.1336 

76 

0 

15 

.24615 

.96923 

.25397 

3.93751 

1.0317 

4.0625 

45 

30 

.25038 

.96815 

.  25862 

3.86671 

1.0329 

3.9939 

30 

45 

.25460 

.96705 

.26328 

3.79827 

1.0341 

3.9277 

15 

15 

0 

.  25882 

.96593 

.26795 

3.73205 

1.0353 

3.8637 

75 

0 

15 

.26303 

.96479 

.27263 

3.66796 

1.0365 

3.8018 

45 

30 

.26724 

.96363 

.  27732 

3  .  60588 

1.0377 

3  .  7420 

30 

45 

.27144 

.96246 

.28203 

3.54573 

1.0390 

3.6840 

15 

16 

0 

.  27564 

.96126 

.  28675 

3.48741 

1.0403 

3.6280 

74 

0 

15 

.27983 

.96005 

.29147 

3.43084 

1.0416 

3.5736 

45 

30 

.  28402 

.95882 

.29621 

3.37594 

1.0429 

3.5209 

30 

45 

.28820 

.95757 

.30097 

3.32264 

1.0443 

3.4699 

15 

17 

0 

.29237 

.95630 

.30573 

3.27085 

1.0457 

3.4203 

73 

0 

15 

.  29654 

.95502 

.31051 

3  .  22053 

1.0471 

3.3722 

45 

30 

.30071 

.  95372 

.31530 

3.17159 

1.0485 

3.3255 

30 

45 

.30486 

.95240 

.32010 

3.12400 

1.0500 

3.2801 

15 

18 

0 

.30902 

.95106 

.32492 

3.07768 

1.0515 

3.2361 

72 

0 

15 

.31316 

.94970 

.32975 

3.03260 

1.0530 

3  .  1932 

45 

30 

.31730 

.94832 

.33460 

2.99868 

1.0545 

3.1515 

30 

45 

.32144 

.94693 

.33945 

2.94591 

1.0560 

3.1110 

15 

19 

0 

.32557 

.94552 

.34433 

2.90421 

1.0576 

3.0715 

71 

0 

15 

.32969 

.94409 

.34922 

2  .  86356 

1.0592 

3.0331 

45 

30 

.33381 

.94264 

.35412 

2.82391 

1.0608 

2.9957 

30 

45 

.33792 

.94118 

.35904 

2.78523 

1.0625 

2.9593 

15 

20 

0 

.34202 

.93969 

.36397 

2.74748 

1.0642 

2.9238 

70 

0 

Cosines 

Sines 

Cotangents 

Tangents 

Cosecants 

Secants 

D 

M 

From  70°  to  80°  read  from  bottom  of  table  upwards. 


SHOP  MATHEMATICS  191 

NATURAL  TRIGONOMETRICAL  FUNCTIONS 


D 

M 

Sines 

Cosines 

Tangents 

Cotangents 

Secants 

Cosecants 

20 

0 

.34202 

.93969 

.36397 

2.74748 

1.0642 

2.9238 

70 

0 

15 

.34612 

.93819 

.36892 

2.71062 

1.0659 

2.8892 

45 

30 

.35021 

.93667 

.37388 

2.67462 

1.0676 

2.8554 

30 

45 

.35429 

.93514 

.37887 

2.63945 

1.0694 

2.8225 

15 

21 

0 

.35837 

.93358 

.38386 

2.60509 

1.0711 

2.7904 

69 

0 

15 

.36244 

.93201 

.38888 

2.57150 

1.0729 

2.7591 

45 

30 

.36650 

.93042 

.39391 

2.53865 

1.0748 

2.7285 

30 

45 

.37056 

.92881 

.39896 

2.50652 

1.0766 

2.6986 

15 

22 

0 

.37461 

.92718 

.40403 

2.47509 

1.0785 

2.6695 

68 

0 

15 

.37865 

.92554 

.40911 

2.44433 

1.0804 

2.6410 

45 

30 

.38268 

.92388 

.41421 

2.41421 

1.0824 

2.6131 

30 

45 

.38671 

.92220 

.41933 

2.38473 

1.0844 

2.5859 

15 

23 

0 

.39073 

.92050 

.42447 

2.35585 

1.0864 

2.5593 

67 

0 

15 

.39474 

.91879 

.42963 

2.32756 

1.0884 

2.5333 

45 

30 

.39875 

.91706 

.43481 

2.29984 

1.0904 

2.5078 

30 

45 

.40275 

.91531 

.44001 

2.27267 

1.0925 

2.4829 

15 

24 

0 

.40674 

.91355 

.44523 

2.24604 

1.0946 

2.4586 

66 

0 

15 

.41072 

.91176 

.45047 

2.21992 

1.0968 

2.4348 

45 

30 

.41469 

.90996 

.45573 

2.19430 

1.0989 

2.4114 

30 

45 

.41866 

.90814 

.46101 

2.16917 

1.1011 

2.3886 

15 

25 

0 

.42262 

.90631 

.46631 

2.14451 

1  .  1034 

2.3662 

65 

0 

15 

.42657 

.90446 

.47163 

2.12030 

1  .  1056 

2.3443 

45 

30 

.43051 

.90259 

.47698 

2.09654 

1  .  1079 

2.3228 

30 

45 

.43445 

.90070 

.48234 

2.07321 

1.1102 

2.3018 

15 

26 

0 

.43837 

.89879 

.48773 

2.05030 

.1126 

2.2812 

64 

0 

15 

.44229 

.89687 

.49315 

2.02780 

.1150 

2.2610 

45 

30 

.44620 

.89493 

.49858 

2.00569 

.1174 

2.2412 

30 

45 

.45010 

.89298 

.50404 

1.98396 

.1198 

2.2217 

15 

27 

0 

.45399 

.89101 

.  50953 

1.96261 

.1223 

2  .  2027 

63 

0 

15 

.45787 

.88902 

.51503 

1.94162 

.1248 

2.1840 

45 

30 

.46175 

.88701 

.52057 

1.92098 

.1274 

2.1657 

30 

45 

.46561 

.88499 

.52613 

1.90069 

.1300 

2.1477 

15 

28 

0 

.46947 

.88295 

.53171 

1.88073 

.1326 

2  .  1300 

62 

0 

15 

.47332 

.88089 

.53732 

1.86109 

1  .  1352 

2.1127 

45 

30 

.47716 

.87882 

.54296 

1.84177 

1.1379 

2.0957 

30 

45 

.48099 

.87673 

.  54862 

1.82276 

1  .  1406 

2.0790 

15 

29 

0 

.48481 

.87462 

.55431 

1.80405 

1  .  1433 

2.0627 

61 

0 

15 

.48862 

.87250 

.56003 

1.78563 

1.1461 

2.0466 

45 

30 

.49242 

.87036 

.56577 

1.76749 

1  .  1490 

2.0308 

30 

45 

.49622 

.86820 

.57155 

1  .  74964 

1.1518 

2.0152 

15 

30 

0 

.50000 

.86603 

.57735 

1.73205 

1  .  1547 

2.0000 

60 

0 

Cosines 

Sines 

Cotangents 

Tangents 

Cosecants 

Secants 

D 

M 

From  60°  to  70°  read  from  bottom  of  table  upwards. 


192  SHOP  MATHEMATICS 

NATURAL  TRIGONOMETRICAL  FUNCTIONS 


D 

M 

Sines 

Cosines 

Tangents 

Cotangents 

Secants 

Cosecants 

30 

0 

.50000 

.86603 

.57735 

.73205 

1.1547 

2.0000 

60 

0 

15 

.50377 

.86384 

.58318 

.71473 

1.1576 

1.9850 

45 

30 

.50754 

.86163 

.58904 

.69766 

1.1606 

1.9703 

30 

45 

.51129 

.85941 

.59494 

.68085 

1.1636 

1.9558 

15 

31 

0 

.51504 

.85717 

.60086 

.66428 

1.1666 

1.9416 

59 

0 

15 

.51877 

.85491 

.60681 

.64795 

1.1697 

1.9276 

45 

30 

.52250 

.85264 

.61280 

.63185 

1.1728 

1.9139 

30 

45 

.52621 

.85035 

.61882 

.61598 

1.1760 

1.9004 

15 

32 

0 

.52992 

.84805 

.62487 

.60033 

1.1792 

1.8871 

58 

0 

15 

.53361 

.84573 

.63095 

1.58490 

1  .  1824 

1.8740 

45 

30 

.53730 

.84339 

.63707 

1.56969 

1.1857 

1.8612 

30 

45 

.54097 

.84104 

.64322 

1.55467 

1.1890 

1.8485 

15 

33 

0 

.54464 

.83867 

.64941 

1.53986 

1.1924 

1.8361 

57 

0 

15 

.54829 

.83629 

.65563 

1.52525 

1.1958 

1.8238 

45 

30 

.55194 

.83389 

.66188 

1.51084 

1.1992 

1.8118 

30 

45 

.55557 

.83147 

.66818 

1.49661 

1.2027 

1.7999 

15 

34 

0 

.55919 

.82904 

.67451 

1.48256 

1.2062 

1.7883 

56 

0 

15 

.  562SO 

.82659 

.68087 

1.46870 

1.2098 

1.7768 

45 

30 

.56641 

.82413 

.68728 

1.45501 

1.2134 

1.7655 

30 

45 

.57000 

.82165 

.69372 

1.44149 

1.2171 

1.7544 

15 

35 

0 

.57358 

.81915 

.70021 

1.42815 

1.2208 

1  .  7434 

55 

0 

15 

.57715 

.81664 

.70673 

1.41497 

1.2245 

1.7327 

45 

30 

.58070 

.81412 

.71329 

1.40195 

1.2283 

1.7220 

30 

45 

.58425 

.81157 

.71990 

1.38909 

1.2322 

1.7116 

15 

36 

0 

.58779 

.80902 

.72654 

1  .37638 

1.2361 

1.7013 

54 

0 

15 

.59131 

.80644 

.73323 

1.36383 

1.2400 

1.6912 

45 

30 

.59482 

.80386 

.73996 

1.35142 

1.2440 

1.6812 

30 

45 

.59832 

.80125 

.74673 

1.33916 

1.2480 

1.6713 

15 

37 

0 

.60181 

.79864 

.75355 

1.32704 

1.2521 

1.6616 

53 

0 

15 

.60529 

.79600 

.76042 

1.31507 

1.2563 

1.6521 

45 

30 

.60876 

.79335 

.76733 

1.30323 

1.2605 

1.6427 

30 

45 

.61222 

.79069 

.77428 

1.29152 

1.2647 

1.6334 

15 

38 

0 

.61566 

.78801 

.78129 

1.27994 

1.2690 

1.6243 

52 

0 

15 

.61909 

.78532 

.78834 

1.26849 

1.2734 

1.6153 

45 

30 

.62251 

.78261 

.79543 

1.25717 

1.2778 

1.6064 

30 

45 

.62592 

.  77;-'8< 

.80258 

1.24597 

1.2822 

1.5976 

15 

39 

0 

.62932 

.77715 

.80978 

1.23490 

1.2868 

1.5890 

51 

0 

15 

.63271 

.77439 

.81703 

1.22394 

1.2913 

1.5805 

45 

30 

.63608 

.77162 

.82434 

1.21310 

1.2960 

1.5721 

30 

45 

.  63944 

.76884 

.83169 

1.20237 

1.3007 

1.5639 

15 

40 

0 

.64279 

.76604 

.83910 

1  .  19175 

1.3054 

1.5557 

.50 

0 

Cosines 

Sines 

Cotangents 

Tangents 

Cosecants 

Secants 

D 

M 

From  50°  to  60°  read  from  bottom  of  table  upwards. 


SHOP  MATHEMATICS  193 

NATURAL  TRIGONOMETRICAL  FUNCTIONS 


M 

Sines 

Cosines 

Tangents 

Cotangents 

Secants 

Cosecants 

40 

0 

.64279 

.  76604 

.83910 

1.19175 

1.3054 

1.5557 

50 

0 

15 

.64612 

.76323 

.84656 

1.18125 

1.3102 

1.5477 

45 

30 

.64945 

.  76041 

.85408 

1  .  17085 

1.3151 

1.5398 

30 

45 

.65278 

.75756 

.86165 

1.16056 

1.3200 

1.5320 

15 

41 

0 

.65606 

.  75471 

.86929 

1  .  15037 

1.3250 

1.5242 

49 

0 

15 

.  65935 

.75184 

.87698 

1  .  14028 

1.3301 

1.5166 

45 

30 

.66262 

.  74896 

.88472 

1.13029 

1.3352 

1.5092 

30 

45 

.66588 

.74606 

.89253 

1.12041 

1.3404 

1.5018 

15 

42 

0 

.66913 

.74314 

.90040 

1.11061 

1.3456 

1.4945 

48 

0 

15 

.67237 

.74022 

.90834 

1  .  10091 

1.3509 

1.4873 

45 

30 

.  67559 

.  73728 

.91633 

1.09131    |     1.3563 

1.4802 

30 

45 

.67880       .73432 

.92439 

1.08179 

1.3618 

1.4732 

15 

43 

0 

.68200   :    .73135 

.93251 

1.07237         1.3673 

1.4663 

47 

0 

15 

.68518       .72837 

.94071 

1.06303         1.3729 

1.4595 

45 

30 

.68835   j    .72537 

.94896 

1.05378 

1.3786 

1.4527 

30 

45 

.69151 

.  72236 

.95729 

1.04461 

1.3843 

1.4461 

15 

44 

0 

.  69466 

.71934 

.96569 

1.03553 

1.3902 

1.4396 

46 

0 

15 

.69779 

.71630 

.97416 

1.02653         1.3961 

1.4331 

45 

30 

.70091 

.71325 

.98270 

1.01761         1.4020 

1.4267 

30 

45 

.70401 

.71019 

.99131 

1.00876         1.4081 

1.4204 

15 

45 

0 

.70711 

.70711 

1.00000 

1.00000 

1.4142 

1.4142 

45 

0 

Cosines 

Sines 

Cotangents 

Tangents     Cosecants 

Secants 

D 

M 

From  45°  to  50°  read  from  bottom  of  table  upwards. 


FORMULAS 


volts  X  amperes      _.     ,  .    . 
H.P.  =  -         -  .     Electric  horse  power. 


Wv2 

K  =  —  —  .     Kinetic  energy  .........  3 

PXPa=WxWa.     Law  of  lever       .      .      .      ,      .      .  8 

PXcos  A  =  WXsin  A.     The  moving  strut.        ...  9 

PXcos  A  =  WX£  sin  A.     The  toggle  joint.       ...  9 


ti 

P=  -  „  —  -.     The  compound  lever.         .       10 

PaXP^XPa^ 

PXR=WXr.    The  wheel  and  axle  ......       10 

-.     The  differential  windlass     ...       11 


P=  —75  —  ^—     5-^.     The  law  for  trains  of  wheels  and 
K  X  -ft/!  X  /v2 

axles  .............       12 


t,     _,     .        .       ,  .  ,  .  .. 

n  =  —  „     „    '    —  -.     The  law  for  driver  and  follower 

r  X-T  j 


pulleys  and  gears  ..........       12 

=  PXN.     The  pulley  block  ........       15 


P=—          —  .     The  differential  pulley  .....       16 
K 

Px  L 

W—  —  77  —  .     The  inclined  plane  when  P  is  parallel  to 
n 

plane  .............       23 


SHOP  MATHEMATICS  195 

Page 

W  =  —  77—.     The  inclined  plane  when  P  is  parallel  to 
n 

base.        .     ...........       23 

P  X  cos  Y 

W  '  =  -  :  -  .     The  inclined  plane  when  P  is  at  an 
sin  a: 

angle  to  incline  ..........       24 

rR 

.    The  screw.          .......       25 


j 
Li 

,    The  wedge  ..........       26 


S=T—-~.     Tap  size  drill  for  V  threads.      ...       29 
N 

S=  71—  ^f.    Tap  size  drill  for  U.  S.  S.  threads.  30 

N 

CXR.  P.  M.Xb      _ 
H.  P.  =  —  —  .     The  horse  power  for  pulleys.      48 

oOO 
p 
/=  —  .     The  coefficient  of  friction  .......       53 


"•  p-=     u  °^  x/x  .v  x  d  x  .ooooos. 

Axle  friction  ...........       54 

W  =  F(1  +0.9  +0.92  +0.93  +0.94).    The  efficiency  of  the 

pulley      ............       56 

sin  A+(fXcosA)  .     .. 

p  =  W  X  -  tt  ,   •  —  TT  •    The  inclined  plane  parallel 

cos  A  —  (fXsin  A) 

to  base.         ...........       57 

g-  The  S(5uare  thread 

R 
screw  .............       58 

thread  so,..      58 


196  SHOP  MATHEMATICS 


.     Length  of  belt.    '  ......       62 


<& 

,     H.  P.  X  50,000          . 

b  =  —  j—     —  £7  —  .     Width  of  belt  .......       64 

dXwXN 

D     3X33,OOOXH.P      _ 

r  =  —       —  y  —       —  .     Pressure  on  bearings.    ...       64 

H.  P.  =  d2xVx.0025.     Horse  power  transmitted  by 

ropes  ...........    *  .  68 

H.  P.  =  dzx  Vx.055.     Horse  power   transmitted    by 

wire  cables  ............       70 

A 

P  D  =  —  —  =.     Diameter  of  chain  sprockets.  72 

sin  1 


.     Diameter  of  shafting 73 

L=6\/d.     Length  of  journal  bearings.     .....       76 

&  =  —.     t  =  —.     Size  of  machine  keys 82 

4  3 

x  =  L  X  -T?-     The  micrometer  screw.  87 

N 

( — - —  I .     Offset      of      footstock       center      for 
I  turning  tapers 101 

C  =  —  — .     The  cutting  speed  in  F.  P.  M.  for  lathe 

turning.         104 

R  =  j-.     Simple  indexing,  milling  machine.        .      .      .     114 

R       F 

—  =  —.     Gearing  for  spiral  head 124 


SHOP  MATHEMATICS  197 


x 
R.P.M.  =  jr.     Speeds  for  drills.      .      .  .      .      129 

WH 
F=^.     Hammer  blow.       .      .  .130 

WXPXR.  P.  M.X*d 

°f 


-'  ^  33,000X12  ' 

machines  .........      ...      132 

H.  P.  =  CxW.     Horse  power  of  machines.        .      .      .      132 

H.  P.  =  —    „„„.     H.  P.  measurement    by    the   dyna- 
00,000 

mometer  ............      136 

F=WXRXN2X.  000341  .  Centrifugal  force  in  fly 

wheels  .............      138 

5  =  W  XRXN2X.  00005427.  The  tensile  strain  in  fly 

wheels.  139 


LJ 

H 
H 
B 

ir 

H 
P 
H 

n 

vN 

iOV 

.      141 
.      145 
.      154 
.      158 

.      160 
.      161 
.      165 

1A7 

oo  ,000 

D           D*N                                                    ^                                          * 

<4.t? 

2t  T  c 

a 

a 
p 

.434 
=  H  A  W     Water  pressure  in  tanks. 
WxH 

•  P-     QQ  ooo'     PumPlng  watci  to  a  given  height. 

V  in  F.  P.  M. 


ANSWERS 


Mechanics,  Page  3 


1.  1,200  ft.-lbs. 

2.  7,500  ft.-lbs. 

3.  80,000  ft.-lbs. 

4.  2,083,350  ft.-lbs. 

5.  3,150  ft.-lbs. 

6.  630  ft.-lbs. 

7.  18,000  ft.-lbs. 

8.  1\  ft.-lbs. 

9.  2,800  ft.-lbs. 

10.  139.9  ft.-lbs. 

11.  10.52  H.  P. 

12.  75.76  H.  P. 


17. 


13.  1.06  H.  P. 

14.  1.35  H.  P. 

15.  140,000  ft.-lbs. 

16.  7,500,000  ft.-lbs. 

V=400ft.  per  sec. 
K  =  30,000  ft.-lbs. 

18.  14.75  H.  P. 

19.  22.12  H.  P. 

20.  10.05  H.  P. 

21.  18.43  H.  P. 

22.  200.38  H.  P. 


Miscellaneous  Problems,  Page  16 


1.  125  Ibs. 

2.  825  Ibs. 

3.  900  Ibs. 

4.  5£  in. 

5.  25.117  in. 

6.  1,024  Ibs. 

7.  164^  Ibs. 

8.  264.49  Ibs. 

9.  1,925.86  Ibs. 

10.  1,417.25  Ibs. 

11.  2,115.94  Ibs. 

12.  8,258.96  Ibs. 

13.  0.0004  in. 


14.  0.0442  in. 

15.  373£lbs. 

16.  3,937^  Ibs. 

17.  1,200  Ibs. 

18.  4,500  Ibs. 

19.  25  in. 

20.  222f  Ibs. 

21.  30  Ibs. 

22.  0.026  dia.  circle 

23.  12^  in. 

24.  4.8  in. 

25.  27|  in. 

26.  125  Ibs. 


SHOP  MATHEMATICS 


199 


27.  1.26  in. 

28.  1|  in. 

29.  SS^lbs. 

30.  390|   Ibs. 

31.  A  ft. 

32.  15f 

33.  24,000  R.  P.  M. 

34.  25  in.  dia. 

35.  ft  rev- 

36.  TV  rev. 

37.  AW  rev. 

38.  72 

39.  63f 

40.  84 


41.  68$ 

42.  93£ 

43.  81f 

44.  69£ 

45.  82f 

46.  4  sheaves 

47.  75  Ibs. 

48.  120  Ibs. 

49.  1,020  Ibs. 

50.  1,105  Ibs. 

51.  92.2% 

52.  150  Ibs. 

53.  4  men 

54.  10  in.  and  9f  in. 


Problems,  Page  27 


1.  118.477  Ibs. 

2.  464.84  Ibs. 

3.  1,118.3  Ibs. 

4.  45,239.04  Ibs. 

5.  71.62  Ibs. 

6.  150.8  Ibs. 

7.  16$  Ibs. 

8.  5,000  Ibs. 

9.  750  Ibs. 
10.  333^  Ibs. 


12. 


11.  29.47  Ibs. 
2}  ft.  ) 
2f  ft.  / 

13.  5 

14.  9 

15.  437  \  Ibs. 

16.  6.631  in. 

17.  7.957  in. 

18.  644.33  Ibs. 

19.  900.68  Ibs. 


3  to  4 


Screw  Threads,  Page  30 

1.  0.0722  in.  5.  0.8375  in. 

2.  0.0481  in.  6.  0.1634  in. 

3.  0.0542  in.  7.  0.3933  in. 

4.  0.1164  in.  8.  0.0139  in. 


200 


SHOP  MATHEMATICS 


9.  0.5913  in. 

10.  0.5068  in. 

11.  0.622  in. 

12.  0.3668  in. 


13.  0.0813  in. 

14.  0.0866  in. 

15.  16  pitch  double 
17.  12  pitch  triple 


Gears,  Page  34 


1.  0.3927  in. 

2.  0.5236  in. 

3.  4  pitch 

4.  10  pitch 

5.  18^  in. 

6.  4  pitch 

7.  108  teeth 

8.  75  teeth 

9.  0.3927  in. 

10.  0.3925  in. 

11.  0.5392  in. 

12.  0.3595  in. 

13.  2i  in. 

14.  3  in. 

15.  4  pitch 

16.  128  teeth 

17.  640  teeth 

18.  200  teeth 

8  in. 
16  in. 
8  in. 
16  in. 
21  in. 
35  in. 


19. 


20. 


21. 


22. 


23. 


24. 


24  in. 
30  in. 
32  in. 
40  in. 
24  in. 
36  in. 

25.  0.0393  in. 

26.  0.0098  in. 

27.  18  in. 

28.  6.6759  in. 

29.  4£  in. 

30.  10  in. 

31.  21.008  R.  P.  M. 

32.  4166J  R.  P.  M. 

132  F.  P.  M. 
.  157.08  F.  P.  M. 

34.  125  Ibs. 

35.  166j 

36.  13.368  R.  P.  M. 

37.  50.93 

38.  3H  Ibs. 

39.  176.84  R,  P.  M. 


33. 


SHOP  MATHEMATICS 


201 


Bevel  Gears,  Page  40 


D 


O  D 


40 
60 

6|  in. 
10  in. 

6.944  in. 
.  10.185  in. 
Cutting  Angles  ( 31°  51'      , 
Index  head        (  54°  29' 

!33°  41' 
56°  19' 

Increment  1°  35' 
Decrement  1°  50' 
Center  face         (70°  32' 
angles  of  blank  \  115°  48' 


N  40 

D4in. 

O  D  4.1414  in. 

Cutting  angle  to  f 
set  head  I 

Angle  C  45° 

Center  face 
angle  of  blank 

Increment  2°  1' 

Decrement  2°  15' 


45' 


94°  2' 


X 


D 


96 
120 
12 
15 

Cutting  Angles  to  (  37°  49' 
set  index  head  \  50°  29' 


2Q, 


,  P 
Angles  G!  and  C  <  5 

Increment  0°  45' 
Decrement  0°  51' 
Center  face          (77°    4' 
angles  of  blank  )  104°  30' 


o 


54 

5    in. 
6§in. 


28°  43' 
56°  47' 


f  4.234  in. 
'  \  6.795  in. 

Cutting  angles  4 

{30°  58' 
59°    2' 

Center  face         f  65°  37' 
angles  blank  1 121°  45' 
Decrement  2°  15' 
Increment  1°  50V 


202 


SHOP  MATHEMATICS 


X 


D 


O  D 


5. 


30 
40 

5    in. 
6|  in. 

5.267  in. 
6.867  in. 

'34°    7' 
50°  23' 
36°  52' 
53°  8' 

Center  face  f  78°  20' 

angles  of  blank  1 110°  52' 
Increment  2°  18' 
Decrement  2°  45' 


D 


O  D 


Cutting  angles 
Angles  G!  and  C 


6. 


fNJ12° 
1 150 

15    in. 
18f  in. 

15.195  in. 
18.906  in. 

38°    0' 
50°  40' 

38°  40' 
51°  20' 

Center  face  f  78°  30' 

angles  of  blank  \  103°  50' 

Increment  0°  35' 

Decrement  0°  40' 


Cutting  angles 
Angles  G!  and  C 


1.  18,000  Ibs. 

2.  11£  in. 

3.  4f  in. 


Worm  Wheels,  Page  42 

4.  47,250  Ibs. 

5.  189,000  Ibs. 


Helical  Gears,  Page  46 


1. 


2, 


3. 


D  3.677  in. 
O  D  4.077  in. 
D  1.414  in. 
0  D  1.614  in. 
D  2.121  in. 
O  D  2.371  in. 


/Centers  2.357  in. 
\O  D  2.524  in. 

5.  1.414  in. 


6. 


No  teeth  11  and  22 
D  for  both  gears  1.754 


Pulleys,  Page  49 


1.  2,040  R.  P.  M. 

2.  12,600  R.  P.  M. 

3.  0.167  in. 

4.  25f  in. 


5.  40  in. 

6.  3.825  in. 

7.  8.325  in. 

8.  11.156  in. 


SHOP  MATHEMATICS 


203 


ft  0.205  in. 

'N6 

9. 

J  c  0.125  in. 

h  4.332  in. 

[T  0.535  in. 

e    1.733  in. 

10. 

3.287  in. 

11 

.   •< 

t    0.5625  in. 

fb  13.82  in. 

T  1.375  in. 

11. 

J  c   0.250  in. 

Dt  8  in. 

[s   0.375  in. 

.L  10.667 

Miscellaneous  Problems,  Page  49 

flead     11.107  in.                   7 

.  75,398.4  Ibs. 

1. 

J  D  3.535  in. 

8.  150,796.8  Ibs. 

lO  D  3.635  in. 

9 

.  75,398.4  Ibs. 

2. 

flead     13.329  in. 
J  D  4.243  in. 
[  O  D  4.443  in. 

10 
11 
12 
13 

.  628,320  Ibs. 
.  6.615  Ibs. 
.  203,575.68  Ibs. 
.  339,292.8  Ibs. 

flead    8.650  in. 

14 

.  73.68  Ibs. 

3. 

J  D  2.946  in. 

15 

.  63,617.4  Ibs. 

[0  D  3.113  in. 

16 

.  49.7  Ibs. 

4. 

16,000  Ibs. 

17 

.  18.65  Ibs. 

5. 

20,000  Ibs. 

18 

.  848,232  Ibs. 

6. 

64,000  Ibs. 

19 

.  37.30  Ibs. 

Friction,  Page 

55 

1. 

2.842  H.  P.                          |     2 

.  13.271  H.  P. 

Miscellaneous  Problems,  Page  58 

1. 

488.39  Ibs. 

5 

.  0 

.08 

2. 

541.18  Ibs. 

6 

.  120  Ibs. 

3. 

474.27  Ibs. 

7 

.  853.85  Ibs. 

4. 

0.125 

8 

.  584.96  Ibs. 

204 


SHOP  MATHEMATICS 


9.  5.6  H.  P. 

10.  148f  Ibs. 

11.  542.87  Ibs. 

12.  92.67  Ibs. 

13.  79.09  Ibs. 

14.  209.5  Ibs. 

15.  2,282.95  Ibs. 


16.  16.76  Ibs. 

17.  33.36  Ibs. 

18.  8,948.02  Ibs. 

19.  1,300.52  Ibs. 

20.  1.7  H.  P. 

21.  8.606  H.  P. 

22.  543.235  ft.-lbs.  per  min. 


23.  360  ft.-lbs.  per  min. 


Belting,  Page  64 


1.  100  H.  P. 

2.  81.92  H.  P. 

3.  96.77  H.  P. 

4.  34.996  ft. 

5.  43.764  ft. 

6.  17^  H.  P. 

7.  20.36  H.  P. 

8.  125.61  H.  P. 

9.  91.48  H.  P. 

10.  139.58  Ibs. 

11.  69.9ft. 

12.  41.5  ft. 

13.  49.3  ft. 

14.  49.25 

15.  24.3  in. 

16.  192  H.  P. 

17.  163.84  H.  P. 

18.  6,600  Ibs. 

19.  756.3  Ibs. 


20.  8.25  Ibs. 

21.  247.5  Ibs. 

22.  66f  in. 

23.  71.42  H.  P. 

24.  408.23  H.  P. 

25.  54.1  H.  P. 

26.  112.64  H.  P. 

27.  4.4  in. 

28.  43.4  in. 

29.  62  \  in. 

30.  20+  oz. 

31.  26+  oz. 

32.  258  R.  P.  M. 

33.  95  R.  P.  M. 

34.  11 1.5  in.  dia. 

35.  52  in. 

36.  30  oz. 

37.  34  oz. 

38.  15.8  oz. 


SHOP  MATHEMATICS 


205 


1.  187ilbs. 

2.  0.675  Ibs. 

3.  5,301.45  Ibs. 

4.  2,650.725  Ibs. 

5.  7,215.86  Ibs. 

6.  14.73  H.  P. 


Rope  Drives,  Page  68 

7.  105.53  Ibs. 

8.  1,100  Ibs. 

9.  42.11  H.  P. 

f  23.11  H.  P. 
10.   J  205  +    R.  P.  M. 

[  65       in. 
11.  1,804  Ibs. 


1. 


1,005  Ibs. 

158.34  H. 
2.  43.53  H.  P. 
399  Ibs. 

54  H.  P. 


'•I3 


Wire  Cables,  Page  70 


P. 


4.  205.19  H.  P. 

5.  2,873.15  F.  P.  M. 

7.41  ft.  dia. 
1,110ft. 


6. 


1. 


2. 


3. 


4. 


o. 


P  D  2.565  in. 
0  D  2.890  in. 
O  D  6.699  in. 
B  D  6.049  in. 
A  0.9  in. 
B  0.6  in. 
b  0.4875  in. 
9.561  in. 
10.049  in. 
B  D  10.974  in. 
P  D  11.461  in. 
O  D  11.949  in. 


fpD 

\0  D 


Chains,  Page  72 

I A  1.05  in. 
6.    <  B  0.7    in. 
[b  0.569  in. 

IB  D  15.038  in. 
P  D  15.607  in. 
O  D  16.176  in. 
(  P  D  8.917  in. 
I  O  D  9.242  in. 
9.  9.229  in. 


206 


SHOP  MATHEMATICS 


Shafting,  Page  74 

d3N           HX80 

10. 

It  in. 

80  '              cl3 

11. 

5.4  in. 

2.  120  H.  P. 

12. 

13.05  in. 

3.  39  H.  P. 

13. 

4iin. 

4.  429.7  H.  P. 

14. 

736  H.  P. 

5.  6.4  in. 

15. 

405  H.  P. 

6.  6i  in. 

16. 

3,645  H.  P. 

7.  46+  R.  P.  M. 

17. 

31+  R.  P.  M. 

8.  4.3  in. 

18. 

592  +  R.  P.  M 

9.  192+  R.  P.  M. 

19. 

308  +  R.  P.  M 

Jack  Shaft,  Page  76 

1.  4.93  in. 

6. 

16|  H.  P. 

2.  3.92  in. 

7. 

22.14  H.  P. 

3.  277.8  R.  P.  M. 

8. 

7.1  in. 

4.  1,312.5  R.  P.  M. 

9. 

10ft. 

5.  450  H.  P. 

10. 

10ft. 

Journal  Bearings, 

Page  79 

1.  1.226  in. 

12. 

13.991  in. 

2.  1.462  in. 

13. 

15.223  in. 

3.  1.087  in. 

14. 

7.111  in. 

4.  7.306  in. 

15. 

2.507  in. 

5.  1.515  in. 

16. 

11.567  in. 

6.  2.915  in. 

17. 

12.219  in. 

7.  9.367  in. 

18. 

3  pulleys 

8.  8.352  in. 

19. 

10.727  in. 

9.  10.283  in. 

20. 

16.121  in. 

10.  11.  124  in. 

21. 

5  pulleys 

11.  4  in. 

22. 

8.67  in. 

SHOP  MATHEMATICS 


207 


23.  20  in. 

24.  14.29  in. 

25.  1.82  in. 

26.  0.17  in. 

27.  iin. 

28.  10  balls 


1. 


29.  1.633  in. 

30.  8  balls 

31.  .126  in. 

32.  iin. 

33.  .343  in. 

34.  .109  in. 


Machine  Keys,  Page  82 

,b  1£  &  1£  in. 
1 1  1    &  .833  in. 
i  b  %  in. 
'  t  .333  in. 


3. 


Micrometer,  Page  88 

1.  &in.  6.  | 

2.  TsW  in-  7.  250  divisions 

3.  stfcr  in.  .  8.  T$v  in. 

4.  i  in.  9.  T«5 

5.  f  10.  50  divisions 

11.  75  divisions 


Lathe  Work,  Page  92 


1.  97+R.  P.  M. 

2.  173+  R.  P.  M. 

3.  103+  R.  P.  M. 


4.  571  +  R.  P.  M. 

5.  82+  R.  P.  M. 

6.  529  +  R.  P.  M. 


Back  Gears,  Page  94 


1.  12+  R.  P.  M. 

2.  32+  R.  P.  M. 

3.  27  +  R.  P.  M. 


4.  15+  R.  P.  M. 

5.  13  +  R,  P.  M. 

6.  19+  R.  P.  M. 


208 


SHOP  MATHEMATICS 


1.  Ratio  of  — — 

2.  60  teeth 

3.  25  teeth 

4.  72  teeth 


Screw  Cutting,  Page  97 

5.  66  teeth 

6.  48  teeth 

7.  24  teeth 

8.  64  teeth 


10. 


9.  Ratio  of  f 
Ratio  of 


1.  7|  in. 

2.  1£  in.  per  ft. 

3.  2^  in.  per  ft. 

4.  s\  in.  per  ft. 

5.  I'j-  in.  per  ft. 

6.  0.64  in.  per  ft. 

7.  0.603  in.  per  ft. 

8.  0.706  in.  per  ft. 


Taper  Turning,  Page  102 

9.  0.6  in.  per  ft. 


10.  3i  in. 

11.  0.411  in. 

12.  1.44  in. 

13.  0.283  in. 

14.  0.833  in. 

15.  0.419  in. 

16.  0.1(32  in. 
17.  0.609  in.  per  ft. 


Miscellaneous  Lathe 

1.  30  R.  P.  M. 

2.  80  R.  P.  M. 

3.  16  R.  P.  M. 

4.  426  +  R.  P.  M. 

5.  32  R.  P.  M. 

6.  40  R.  P.  M. 

7.  46+  R.  P.  M. 

8.  256  R.  P.  M. 

9.  80  R.  P.  M. 


Problems,  Page  105 

10.  19+  R.  P.  M. 

11.  10  R.  P.  M. 

12.  61+  R.  P.  M. 

13.  16  R.  P.  M. 

14.  95+  R.  P.  M. 

15.  466  +  R.  P.  M. 

16.  34  R.  P.  M. 

17.  32  R.  P.  M. 

18.  13.716+  in. 


SHOP  MATHEMATICS 


209 


19. 

(  16.266  in. 
(  15.533  in.  approx. 

22.  0.513  in. 
23.  0.389  in. 

20. 

(  26.297  in. 

24.  114+  R.  P.  M. 

(  25.112  in.  approx. 

25.  8  R.  P.  M. 

21. 

10.412  in. 

Planer,  Page  109 

1. 

17.864  F.  P.  M. 

6.  54+  min. 

2. 

46.584  F.  P.  M. 

7.  34+  min. 

3. 

22.317  F.  P.  M. 

8.   V 

4. 

16.788  F.  P.  M. 

n     (  24"  in.  dia. 

5. 

29  +  min. 

9     \ 
(18       in.  dia. 

Milling  Machine,  Page  115 

Simple  Indexing 

1. 

13  1  turns 

10.  i^turn 

2. 

H 

H.  H 

3. 

10 

12.  i? 

4. 

If 

13.  H 

5. 

Ij 

14.  i 

6. 

§ 

15.  A 

7. 

$ 

16.  i 

8. 

H 

17.  A 

.18.  A 
Compound  Indexing,  Page  118 

OfJ       30 30    r\r  94.       13 3      r>r 

£\).     Jy ^^   Ul  ^t.     ^T£ 4^   Ol 

A+A  H— A 

21.  A+H  25.  &—&OT 

22.  A+A  or  H— M 
t»— A  26.  ii— A 

23.  A+A  or  27. 
H-T&  28. 


210 


SHOP  MATHEMATICS 


Differential  Indexing 

So   many   combinations   possible   that   answers*  are   omitted. 


1.  Ratio  of  J 

2.  Ratio  of  y 

3.  Ratio  of  | 

4.  Ratio  of  V 

5.  Ratio  of  V 

6.  9.524  in. 

7.  36  in. 

8.  2i  in. 

9.  21°  26' 
10.  27°  38' 

34°  30' 
18°  36' 
13°  15' 

14.  16.327  in. 

15.  31.416  in. 


11. 
12. 
13. 


16. 


The  Spiral  Head,  Page  125 

, 18°  58' 
6  min. 

17.  41  +  R.  P.  M. 

18.  21  +  R.  P.  M. 

19.  40  R.  P.  M. 

20.  80  R.  P.  M. 

21.  192  R.  P.  M. 

22.  128  R.  P.  M. 

23.  80  R.  P.  M. 

24.  86  +  R.  P.  M. 

25.  18.54  F.  P.  M. 

50.4  F.  P.  M. 
35+  R.  P.  M. 

27.  0.617  in. 

28.  0.278  in. 

29.  1.867  in. 


26. 


1. 


6. 


Drill  Press,  Page  129 

293  R.  P.  M. 
41  sec. 

2.  0.64  in. 

3.  57  sec. 

4.  47  sec. 

52+  sec. 
125+  R.  P.  M. 


5. 


1  min.  22  sec. 
220  R.  P.  M. 

7.  4  min.  12  sec. 

8.  4  min.  4  sec. 

9.  4  min. 


SHOP  MATHEMATICS 


211 


1.  288,000  Ibs. 

[329,143  Ibs. 

2.  1 384,000  Ibs. 
[460,800  Ibs. 

3.  276,000  Ibs. 


Hammer  Blow,  Page  131 

4.  326,400  Ibs. 

5.  20,000  Ibs. 

6.  288,000  Ibs. 

7.  10,000  Ibs. 

8.  20,000  Ibs. 
13,125  Ibs. 


Horse  Power  of  Machines,  Page  133 

1.  0.589  H.  P.  10.  0.551  H.  P. 

2.  0.929  H.  P.  11.  0.506  H.  P. 

3.  0.524  H.  P.  12.  1.875  H.  P. 

4.  0.220  H.  P.  13.  2.142  H.  P. 

5.  0.23    H.  P.  14.  2.076  H.  P. 

6.  1.805  H.  P.  15.  .785  H.  P. 

7.  1.106  H.  P.  16.  .880  H.  P. 

8.  0.413  H.  P.  17.  1.276  H.  P. 

9.  0.743  H.  P.  18.  2.856  H.  P. 


1.  281 +R.  P.  M. 

2.  18.75  H.  P. 


Dynamometers,  Page  136 

3.  25  H.  P. 

4.  2.801  ft. 
5.  50  Ibs. 

Fly  Wheels,  Page  139 


1.  19,432.7  Ibs. 

2.  414.6  Ibs. 

3.  13,063  +  Ibs. 

4.  7,124+  Ibs. 

5.  14,563  +  Ibs. 

6.  9,167+  Ibs. 

7.  77,676  +  Ibs. 


8.  3,607  +  Ibs. 

9.  9,770  +  Ibs. 

10.  210+  R.  P.  M. 

11.  480+  R.  P.  M. 

12.  136+  R.  P.  M. 

13.  5,654.88  F.  P.  M. 

14.  940.368  F.  P.  M. 


212 


SHOP  MATHEMATICS 


Horse  Power  of 

1.  2(5:5.2  H.  P. 

2.  215.0  H.  P. 
:;.   177.4  H.  P. 

4.  719.7  II.  P. 

5.  0.476  H.  P.  porlb. 

M.  E.  P. 

0.  26.00  H.  P.  porlb. 
M.  E.  P. 

7.  1.37  H.  P.  perlb. 

M.  E.  P. 

8.  O.S675  II .  P.  pcrlb. 

M.  E.  P. 

9.  423.1  H.  P. 

10.  415.9  H.  P. 

11.  15,402  Ibs. 

15,360  lb/. 

552.7  1}.  P.  at  1  mile 
[      per  In  in. 

13.  10.95  1'bs.  M.  E.  P. 

14.  24.3  Ibs.  M.  E.  P. 

15.  2.67+  H.  P. 

16.  5£  H.  P. 

17.  1HH.  P. 


Engines,  Page  142 

18.  33.47  H.  P. 

19.  52+  H.  P. 

20.  22 ±  H.  P. 

21.  25.9  H.  P. 

22.  6+  H.  P. 

23.  320.8  H.  P. 

24.  9.996  H.  P.  per  lb. 

M.  E.  P. 

25.  18,488.89  Ibs. 

26.  19,413.33  Ibs. 

27.  18.838  in. 

28.  4.318  H.  P.  per  11 ». 

M.  E.  P. 

29.  10.927  in. 

30.  17.019  in. 

31.  10.249  in. 

32.  9.577  in. 

^  8  in.  crank 
l>182+  R.  P.  M. 

34.  435.4  H.  P. 

35.  4.752  H.  P.  per  lb. 

M.  E.  P. 


1.  $210 

2.  $3,852.90 
::.  10.4S  H.  P. 

4.  Hi5.2  II.  P. 

5.  1899.4  sq.  ft. 


Steam  Boiler,  Page  155 

6.  1386.8  sq.  ft. 

7.  1267.2  sq.  ft. 

8.  169.65  cu.  ft, 

9.  1141.2  sq.  ft. 
10.  40.2  H.  P. 


SHOP  MATHEMATICS 


213 


11. 

36.7  H.  P. 

20.  .589  in. 

12. 

13. 
14. 
15. 
16. 

$20.16 
9J  sq.  ft. 
1,728  gals. 
2,000  gals. 
174.36  cu.  ft. 

(  861+  ibs. 

•    (  1283  +  sq.  ft. 
22.  689  Ibs. 
23.  1,167  Ibs. 
24.   168  Ibs. 

17. 

232.3 

25.   .474  in. 

18. 

312^  Ibs. 

26.   .348  in. 

19. 

00  1.5  Ibs. 
28.  53,  : 

27.  62,261  Ibs. 
,42  Ibs. 

Safety  Valve,  Page  159 

2. 
3. 

149.08  Ibs.                                 4.  45  in. 
6.652+  in.                                5.  175-V  Ibs. 

Hydraulic  Machines,  Page  161 

1. 

11.38                                         10.  44,957.4  Ibs. 

•j. 

.914                                             11.  2  1.3  Ibs. 

4. 

.962 
9.92 

12.  4,000  Ibs. 
13.  90.228  Ibs. 

5. 

86.8  Ibs. 

14.   14.114  Ibs. 

6. 

27(».5  ft. 

15.  245.41  Ibs. 

7. 

1,687.5  Ibs. 

16.  392.65  Ibs. 

8. 
9. 

S43|  Ibs. 
9,375  Ibs. 

17.  98.16  Ibs. 
18.  94.85  ft. 

Steam  Pumps,  Page  166 


1.  8,812.68  gals. 

2.  1.48  H.  P. 

3.  3,304.8  gals. 


4.  22,521  Ibs. 

5.  2;208.9  cu.  ft. 

6.  2.79  H.  P. 


214 


SHOP  MATHEMATICS 


7.  3 

,427.2 

13.  6.632  in. 

8.  29.75  H.  P. 

14.  68.177  cu.  ft. 

9.  2 

.22  H.  P. 

15.  9.29  H.  P. 

10.  37.975  Ibs.  per 

sq.  in. 

16.  10  in.  nearly 

11.  64  min.  37  sec. 

17.  17.7  Ibs. 

12.  3 

.346  in. 

18.  21.8  H.  P. 

19.  12.988  in. 

Formulas,  Page  183 

1.  50 

c             a 

2.  20 

V.       \~>  .             A    1     "  

sm  A 

3.  108 

4.  112 

7.  £  or  .5 

WXb   ,, 

WXb 

9.  25°  C 

-p-;  F 

a 

10.  -7f°  C 

5. 

,     PXa  w 

PXa 

11.  50°  F 

W  ' 

~b~~ 

12.  14°  F 

INDEX 


Abbreviations ix 

Addendum 32 

Adhesion  of  belts 61 

Angle  to  set  milling  ma- 
chine table 126 

Angle  of  friction 55 

Archimedes'  discovery.  .  161 

Arithmetical  signs ix 

Avoirdupois  pound 170 

Axle  friction 53 

B 

Bar  iron,  rule  for  weight 

of 172 

Ball  bearings 77 

Bearings,  belt  pressure. .  64 

Bearings,  journal 76 

Bearings,  shaft 76 

Bearings,  weight  on.  ...  77 

Belts,  leather 61 

Belts,  H.  P.  of 64 

Belts,  length  of 62 

Belts,  centrifugal  force  in  62 

Bevel  gears 37 


Boilers,  steam 149 

Boiler,  heating  surface  of  151 

Boilers,  strength  of 154 

Boilers,  water  capacity. .  152 
Breaking  strain  of  man- 

ila  rope 68 

Breaking  strain,  wire 

rope 70 


Capstan 10 

Centrifugal  force  in  pul- 
ley rims 138 

Chain  transmission 71 

Chain  link 71 

Coefficient  of  friction ...     52 
Combination  of  wheels, 

gears,  examples 49 

Combination  of  screws, 

pulleys,  etc.,  examples     51 
Compound   indexing  on 

milling  machine 1 16 

Compound  lever,  mech. 
power 9 


216 


INDEX 


Cutting  speeds  for  vari- 
ous materials 104 

D 

Decimal  equivalents. . . .  177 

Dedendum 32 

Depth  of  gear  teeth.  ...  34 

Diametral  pitch 32 

Differential  indexing 

milling  machine 119 

Differential  pulley 15 

Differential  screw 25 

Differential  windlass. ...  11 

Drill  press 129 

Driver  and  follower, 

rules  for 12 

Dynamometer 135 

E 

Electric  H.  P 2 

Emery  wheel,  speed  of .  .  36 
Energy,    potential    and 

kinetic 3 

Engine  details 147 

Engine  fly  wheel 138 

Engines,  gas  and  gaso- 
lene   145 

Engine  lathe 91 

F 

Factor  of  safety 70 

Feather  kevs.  .  82 


PAGE 

Feeds,  lathe  carriage ...  90 

Feed,  lead  screw 95 

Fixed  pulley 14 

;  Fly  wheel 138 

;  Force 1 

Formulas,    use    of 178 

Friction 52 

Friction,  laws  for 55 

Friction  in  bearings ....  54 
Friction   in   mechanical 

powers - 55 

G 

Gallon,  cubic  contents. .  170 

Gas  unit  of  measure. ...  160 

Gasolene  engines •  .  145 

Gear  wheel  definition ...  32 

Gears,  rules  for  pitch  of.  33 

Gear,  spiral 42 

Gear,  worm 41 

General  law  of  machines  6 

Grinder,  speeds  for 37 

Grindstone  speed 37 

H 

Hammer  blow 130 

Head  of  water 160 

Helical  gears 42 

Horse  power,  definition .  2 

•  Horsepower  in  belts. ...  62 

in  boilers 150 

in  electrical  machinery  2 

in  the  locomotive. .  144 


INDEX 


217 


in  machines 132 

in  pulleys 48 

in  manila  ropes 68 

in  shafting 73 

in  steam  engines 141 

in  pumps 165 

in  wire  cables 70 

Hydraulics 160 

Hydraulic  machines. .  163 


Inclined  plane  mechani- 
cal power 23 

Inclined  plane,  friction 
in 56 

Inclined  plane,  angle  of 
friction  for 55 

Index  head,  milling 
machine 112 

Indexing,  compound  ...    116 

Indexing,  differential.  ..    llg 


Jack  shaft 75 

Journal  for  ball  bearings  77 
Journal    bearing    allow- 
able pressure  on 77 

Journal  bearing  for  ma- 
chines   77 

Journal  for  shafting.  .  76 


Keys,  feather 82 

Keys,  machine 82 


Lathe,  description  of .  .  .  91 

Lathe,  carriage  feed  for.  90 

Lathe,  change  gears  for.  95 
Lathe,  compound  gears 

for ! 98 

Lathe,  back  gears  for.  .  .  93 

Lathe,  reversing  gear  for  97 

Lathe,  lead  screw  for.  .  .  95 

Lathe,  feed  rod  for 91 

Lathe,  speeds  for  cuts  on  104 

Lathe  test  indicator. ...  18 

Laws  for  friction 55 

Leather  belts 61 

Leather  belts,  lengths  of  62 

Length,  unit  of 169 

Lever,  three  kinds 7 

Lever,  bent 8 

Lever,  calculations  for. .  8 

Liquid  measure 170 

Locomotive,  power  of . . .  143 

If 

Machines 90 

Machines,    general    law 

for. .  6 

,   Machine  keys 82 


218 


INDEX 


Machines,  general 90 

Manila  rope 68 

Materials,  weights  and 

melting  points 172 

Mathematical  signs ix 

Measurement,  units  of .  .  169 
Measurement  by 

micrometer 83 

Measurement  of  liquid .  .  170 

Mechanical  powers 6 

Melting  points  of  metals  172 

Meter 173 

Metric  tables 173 

Micrometer  explained. .  .  83 

Mile  in  feet 169 

Milling  machine Ill 

Milling  cutters 127 

Moment  of  forces 1 

Movable  pulley 14 

N 

Natural  sines,  etc.,  table  181 

O 

Open  belt,  lengths  of. ...     63 
Offsetting  the  center  for 
taper  turning 101 

P 

Planer ' 108 

Power  and  speed,  Gene- 
ral   law 6 

Powers,  mechanical ....        6 


Pressure  in  Ibs.  per  sq. 

in 169 

I  Pressure  or  head  of  water  160 

Pressure  on  bearings  by 

belts 64 

Pressure  on  bearings  by 

weight 54 

Pressure,  unit  of  meas- 
ure   169 

Pulley,     mechanical 

power 13 

Pulley  design 47 

Pulley  safe  speed 47 

Pumps 163 

I 

R 

Reversing  gears  for  lathe  97 

Roll  bearings 77 

Rolling  friction 52 

Rope,  breaking  strain  of  68 

Rope,  manila 68 

Rope  transmission 68 

Rule  for  trains  of  wheels 

or  pulleys 12 

Rules  for  driver  and  fol- 
lower   12 

S 

Safe  speed  for  pulleys.  .  .  47 

Safety*  valves 157 

Screw  cutting,  lathe. ...  95 

Screw,  differential 25 


INDEX 


219 


Screw,  friction  in 58 

Screw,  mechanical  power  24 

Screw,  lead  or  pitch  of .  .  24 

Screw  threads 29 

Shafting ..  73 

Sliding  friction 52 

Speeds  of  pulleys,  safe .  .  47 

Speeds  of  emery  wheels .  36 

Speeds  of  grindstones.  .  .  37 

Speeds  of  lathe  cuts 104 

Speeds  of  drills 129 

Speeds  of  milling  cutters  127 
Speeds  of  polishing 

wheels 37 

Speed  and  power,  gen- 
eral law 6 

Spiral  gears 42 

Spiral  head,  milling  ma- 
chine   123 

Sprockets 71 

Spur  gears 32 

Square  measure 169 

Steam  boiler 149 

Steam  engine  formulas. .  141 

Steam  engine  details. . .  .  147 

Strut,  moving 9 

Symbols,  mathematical .  ix 

T 

Table  of  decimal  equiva- 
lents   177 

Table,  natural  sines,  etc .  181 


Tapers  per  foot 102 

Tapers  per  inch 102 

Taper  turning 101 

Tension  in  pulleys 139 

Trigonometric  functions  186 
Thread  cutting,  lathe ...  95 
Threads,  U.  S.,  V,  Whit- 
worth 29 

Toggle  joint 9 

U 

Units  of  length,  etc 169 

Use  of  formulas 178 

V 

Velocity  of  pulleys 47 

Velocity  ratio 32 

Vernier  caliper 85 

V-threads 29 

W 

Water  pressure 160 

Wedge,    mechanical 

power 26 

Weight  of  water 170 

Weight  of  metals 172 

Windlass,  simple 10 

Windlass,  differential. . .  11 

Wheel  and  axle 10 

Wire  rope,  H.  P.,  etc ...  70 

Wire  rope  sheaves 69 

Whitworth  threads 30 

Work 1 

Worm  gears 41 


A     000  038  664     9 


